Ace Legona

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b+1}{8}+\dfrac{c+1}{8}\)

\(\ge3\sqrt[3]{\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\cdot\dfrac{b+1}{8}\cdot\dfrac{c+1}{8}}=\dfrac{3a}{4}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4};\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\Leftrightarrow VT\ge\dfrac{3}{4}=VP\)

Khi \(a=b=c=1\)

\(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)

ĐK:\(-2\le x\le 2\)

\(\Leftrightarrow4\left(2x+4\right)+16\left(2-x\right)+16\sqrt{2\left(4-x^2\right)}=9x^2+16\)

\(\Leftrightarrow8\left(4-x^2\right)+16\sqrt{2\left(4-x^2\right)}+16=x^2+8x+16\)

\(\Leftrightarrow\left(2\sqrt{2\left(4-x^2\right)}+4\right)^2=\left(x+4\right)^2\)

Vì \(-2\le x\le 2\)\(\Rightarrow\left\{{}\begin{matrix}x+4>0\\2\sqrt{2\left(4-x^2\right)}+4>0\end{matrix}\right.\)

\(\Rightarrow2\sqrt{2\left(4-x^2\right)}+4=x+4\)

\(\Leftrightarrow2\sqrt{2\left(4-x^2\right)}=x\Leftrightarrow8\left(4-x^2\right)=x^2\Rightarrow x=\dfrac{4\sqrt{2}}{3}\)

ĐK:\(x\ge 1;y\ge 1\)

Áp dụng BĐT AM-GM ta có:

\(VT=x\sqrt{y-1}+2y\sqrt{x-1}\)

\(\le x\cdot\dfrac{y-1+1}{2}+2y\dfrac{x-1+1}{2}\)

\(=\dfrac{xy}{2}+\dfrac{2xy}{2}=\dfrac{3xy}{2}=VP\)

Xảy ra khi \(x=y=2\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)

ĐK:\(x\ge 1\)

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\dfrac{x+3}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\)

\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{x-1}+\left|\sqrt{x-1}-1\right|=\dfrac{x+1}{2}\)

*)Xét \(x\le 1\)\(\Rightarrow\sqrt{x-1}-\sqrt{x-1}+1=\dfrac{x+1}{2}\)

\(\Rightarrow\dfrac{x+1}{2}=1\Rightarrow x+1=2\Rightarrow x=1\)

*)Xét \(x>1\)\(\Rightarrow\sqrt{x-1}+\sqrt{x-1}-1=\dfrac{x+1}{2}\)

\(\Rightarrow2\sqrt{x-1}=\dfrac{x+3}{2}\)\(\Rightarrow4\sqrt{x-1}=x+3\)

\(\Rightarrow16\left(x-1\right)=x^2+6x+9\)\(\Rightarrow x=5\)

Vậy \(x=1;x=5\)

Đặt \(AB=a;AC=b;AD=c\). Kẻ \(DE\) vuông góc \(AC(E\in AB; F\in AC)\)

Ta có: Tứ giác \(AFDE\) là hình chữ nhật do \(\widehat{A}=\widehat{E}=\widehat{F}=90^o\), AD phân giác trong của \(\widehat{EAF}\) nên \(\widehat{AFDE\:}\) là hình vuông. Suy ra

\(DE=DF=\dfrac{AD\sqrt{2}}{2}=\dfrac{c\sqrt{2}}{2}\). Ta có:

\(S_{DAB}+S_{DAC}=S_{ABC}\)

\(\Leftrightarrow\dfrac{1}{2}AB\cdot DE+\dfrac{1}{2}DF\cdot AC=\dfrac{1}{2}AC\cdot AB\)

\(\Leftrightarrow\dfrac{c\sqrt{2}}{2}a+\dfrac{c\sqrt{2}}{2}b=ab\)

\(\Leftrightarrow\dfrac{\sqrt{2}}{c}=\dfrac{1}{a}+\dfrac{1}{b}\). Hay \(\dfrac{\sqrt{2}}{AD}=\dfrac{1}{AB}+\dfrac{1}{AC}\)

Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

Và \(a+b+c=\sqrt{2017}\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2017\)

\(\Rightarrow a^2+b^2+c^2+2\cdot0=2017\)\(\Rightarrow a^2+b^2+c^2=2017\)

Vậy \(a^2+b^2+c^2=2017\)

#5k->Acc_Quỳnh,pls :)

Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

Và \(a+b+c=\sqrt{2017}\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2017\)

\(\Rightarrow a^2+b^2+c^2+2=2017\Rightarrow a^2+b^2+c^2=2015\)

Vậy \(a^2+b^2+c^2=2015\)

#5k->Acc_Quỳnh,pls :)

T5/483 Toán tuổi thơ số 483-9/2017, ngại lật báo quá ko biết nó có đáp án k nữa .-.

*)Xét \(ab+bc+ca<0\) hiển nhiên đúng

*)Ta cần chứng minh \(ab+bc+ca\ge 0\)

Đặt \(\left\{{}\begin{matrix}a+b+c=3u\\ab+bc+ca=3v^2\\abc=w^3\end{matrix}\right.\). Ta c/m 1 BĐT tuyến tính của \(w^3\)

Xảy ra các trường hợp sau đây

+)\(w^3=4\) *Đúng*

+)\(b=a;c=3-2a\). Khi đó \(a^2(3-2a)\geq-4\)

\(\Leftrightarrow (a-2)(2a^2+a+2)\leq0\)

Với \(a\le 2\) thì ta cần cm \(3a^2(3-2a)+12\geq5(a^2+2a(3-2a))\)

\(\Leftrightarrow (a-1)^2(2-a)\geq0\)

Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:

\(VT=\sqrt{10-x}+\sqrt{x+6}\)

\(\ge\sqrt{10-x+x+6}=\sqrt{16}=4\)

Và \(VP=-x^2-12x-32=-x^2-12x-36+4\)

\(=-\left(x^2+12x+36\right)+4\)\(=-\left(x+6\right)^2+4\le4\)

\(\Rightarrow VT\ge VP=4\) xảy ra khi \(VT=VP=4\)

\(\Rightarrow-\left(x+6\right)^2+4=4\Rightarrow x=-6\)

Từ \(\dfrac{a}{1+a}+\dfrac{2b}{2+b}+\dfrac{3c}{3+c}\le\dfrac{6}{7}\)

\(\Leftrightarrow1-\dfrac{a}{1+a}+2-\dfrac{2b}{2+b}+3-\dfrac{3c}{3+c}\ge6-\dfrac{6}{7}\)

\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\ge\dfrac{36}{7}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\)

\(\ge\dfrac{\left(1+2+3\right)^2}{a+b+c+6}=\dfrac{36}{7}=VP\)

Xảy ra khi \(a=\dfrac{1}{6};b=\dfrac{1}{3};c=\dfrac{1}{2}\)

Ta có:

\(\sqrt[4]{4}VT=\sqrt[4]{4a^3}+\sqrt[4]{4b^3}+\sqrt[4]{4c^3}\)

\(=\sqrt[4]{\left(a+b+c\right)a^3}+\sqrt[4]{\left(a+b+c\right)b^3}+\sqrt[4]{\left(a+b+c\right)c^3}\)

\(>\sqrt[4]{a^4}+\sqrt[4]{b^4}+\sqrt[4]{c^4}=a+b+c=4\)

\(\Rightarrow\sqrt[4]{4}VT>4\Rightarrow VT>\dfrac{4}{\sqrt[4]{4}}=2\sqrt{2}\)

Ta có: \(A=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}>0\)

\(\Rightarrow A^2=6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}\)

\(\Rightarrow A^2=6+A\)\(\Rightarrow A^2-A-6=0\)

\(\Rightarrow A^2-2A+3A-6=0\)

\(\Rightarrow A\left(A-2\right)+3\left(A-2\right)=0\)

\(\Rightarrow\left(A+3\right)\left(A-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}A+3=0\\A-2=0\end{matrix}\right.\)\(\Rightarrow A=2>0\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{x-4}+\sqrt{6-x}\right)^2\)

\(\le\left(1+1\right)\left(x-4+6-x\right)=4\)

\(\Rightarrow VT^2\le4\Rightarrow VT\le2\left(1\right)\)

Và \(VP=x^2-10x+27=x^2-10x+25+2\)

\(=\left(x-5\right)^2+2\ge2\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow VP\le VT=2\)

Khi \(VP=VT=2\Rightarrow x=5\)

\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Leftrightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

Xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}=2\\\sqrt{y}=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=9\end{matrix}\right.\)

Đặt \(\left(a;b;c\right)\rightarrow\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\) thì:

\(\left(-x+y+z\right)\left(x-y+z\right)\left(x+y-z\right)\le xyz\)

Áp dụng BĐT AM-GM:

\(\sqrt{\left(-x+y+z\right)\left(x-y+z\right)}\le\dfrac{-x+y+z+x-y+z}{2}=z\)

Tương tự rồi cho 2 BĐT còn lại cũng có:

\(\sqrt{\left(x-y+z\right)\left(x+y-z\right)}\le x;\sqrt{\left(-x+y+z\right)\left(x+y-z\right)}\le y\)

Nhân theo vế 3 BĐT trên ta có:

\(VT=\left(-x+y+z\right)\left(x-y+z\right)\left(x+y-z\right)\le xyz=VP\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\Rightarrow\dfrac{1}{8}\ge abc\)

Áp dụng BĐT Holder ta có:

\(B=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)

\(\ge\left(\sqrt[3]{3\cdot3\cdot3}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\right)^3\)

\(=\left(3+2\sqrt[3]{\dfrac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\dfrac{1}{\dfrac{1}{8}}}\right)^3=343\)

Khi \(a=b=c=\dfrac{1}{2}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT=\left(a^2+4\right)\left(b^2+9\right)\)

\(\ge\left(\sqrt{a^2b^2}+\sqrt{4\cdot9}\right)^2=\left(ab+36\right)^2=VP\)

Xảy ra khi \(\dfrac{a^2}{4}=\dfrac{b^2}{9}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow b=\dfrac{3a}{2}\)

Khi đó \(A=\dfrac{a^2-ab+b^2}{a^2+ab+b^2}=\dfrac{a^2-a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}{a^2+a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}=\dfrac{7}{19}\)

\(x^2-2x+y^2+4y+6\)

\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+1\)

Ta thấy: \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Cho \(a,b,c>0\) và \(ab+bc+ca=3\). CMR \((a^ab^bc^c)^{\dfrac{3}{a+b+c}}\ge\sqrt[3]{\dfrac{a^3+b^3+c^3}{3}}\)

#Bài này số mũ đẹp quá nên chưa nghĩ ra cái j cả :v

a)\(4x^4+y^4\)

\(=4x^4+4x^2y^2+y^4-4x^2y^2\)

\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)

\(=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)

b)\(x^5+x^4+1\)

\(=x^5+x^3+x^3-x^3-x^2-x+x^2+x+1\)

\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)

\(\sqrt{x+4\sqrt{x-4}}=3\)

\(\Leftrightarrow\sqrt{x-4+4\sqrt{x-4}+4}=3\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=3\)

\(\Leftrightarrow\sqrt{x-4}+2=3\)

\(\Leftrightarrow\sqrt{x-4}=1\Leftrightarrow x-4=1\Leftrightarrow x=5\)

Khó quá. Đúng là Câu Hỏi Hay!!

a)Áp dụng BĐT AM-GM ta có:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

Nhân theo vế 2 BĐT trên có:

\(A\ge9\sqrt[3]{abc\cdot\dfrac{1}{abc}}=9\)

Khi \(a=b=c\)

Bài 2:

a)Sửa đề \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{\left(1+1\right)^2}{x+y}=\dfrac{4}{x+y}\)

Khi \(x=y\)

b)Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có:

\(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}=\dfrac{4}{2b}=\dfrac{2}{b}\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c};\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\ge\dfrac{2}{a}\)

Cộng theo vế 3 BĐT trên ta có:

\(2VT\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2VP\Leftrightarrow VT\ge VP\)

Khi \(a=b=c\)

#Đêm qua tự nhiên mơ thấy cách này, dậy làm luôn :v

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(x^2+y^2+1\right)\left(1+1+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow\dfrac{1}{x^2+y^2+1}\le\dfrac{2+z^2}{\left(x+y+z\right)^2}.\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\dfrac{1}{y^2+z^2+1}\le\dfrac{2+x^2}{\left(x+y+z\right)^2};\dfrac{1}{x^2+z^2+1}\le\dfrac{2+y^2}{\left(x+y+z\right)^2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\le\dfrac{x^2+y^2+z^2+6}{\left(x+y+z\right)^2}=\dfrac{x^2+y^2+z^2+2\left(xy+yz+xz\right)}{\left(x+y+z\right)}=1\)

Khi \(x=y=z=1\)

Đề bắt quy nạp khó quá, giá đề mở thì xài Ber's ineq cho lẹ .-.

*) Với \(n=1;2\) BĐT đúng

*)Giả sử BĐT đúng với \(n=k\) tức chứng minh BĐT đúng với \(n=k+1\) hay \(\dfrac{a^{k+1}+b^{k+1}}{2}\ge\left(\dfrac{a+b}{2}\right)^{k+1}\)

Ta có: \(VT-VP=\dfrac{a^{k+1}+b^{k+1}}{2}-\left(\dfrac{a+b}{2}\right)^{k+1}\)

\(=\dfrac{a^{k+1}+b^{k+1}}{2}-\left(\dfrac{a+b}{2}\right)^k\left(\dfrac{a+b}{2}\right)\)

\(\ge\dfrac{a^{k+1}+b^{k+1}}{2}-\dfrac{a^k+b^k}{2}\cdot\dfrac{a+b}{2}\)

\(=\dfrac{\left(a-b\right)\left(a^k-b^k\right)}{4}=\dfrac{\left(a-b\right)\left(a^{k-1}-a^{k-2}b+...+b^{k-1}\right)}{4}\ge0\)

Khi \(a=b\)

Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\)\(\left(x,y,z>0\right)\)\(\Rightarrow\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\x+z=2b\end{matrix}\right.\)

Thì ta có: \(\dfrac{2\left(y+z\right)}{x}+\dfrac{9\left(x+z\right)}{2y}+\dfrac{8\left(x+y\right)}{z}\ge26\)

Áp dụng BĐT AM-GM ta có:

\(VT=\dfrac{2\left(y+z\right)}{x}+\dfrac{9\left(x+z\right)}{2y}+\dfrac{8\left(x+y\right)}{z}\)

\(=\dfrac{2y}{x}+\dfrac{2z}{x}+\dfrac{9x}{2y}+\dfrac{9z}{2y}+\dfrac{8x}{z}+\dfrac{8y}{z}\)

\(=\left(\dfrac{2y}{x}+\dfrac{9x}{2y}\right)+\left(\dfrac{2z}{x}+\dfrac{8x}{z}\right)+\left(\dfrac{9z}{2y}+\dfrac{8y}{z}\right)\)

\(\ge2\sqrt{\dfrac{2y}{x}\cdot\dfrac{9x}{2y}}+2\sqrt{\dfrac{2z}{x}\cdot\dfrac{8x}{z}}+2\sqrt{\dfrac{9z}{2y}\cdot\dfrac{8y}{z}}\)

\(\ge6+8+12=26=VP\)

Đặt \(\left\{{}\begin{matrix}\sqrt{3x^2-2x+9}=a\\\sqrt{3x^2-2x+2}=b\end{matrix}\right.\)\(\left(a,b\ge0\right)\) có:

\(\left\{{}\begin{matrix}a+b=7\\a^2-b^2=7\end{matrix}\right.\). Khi đó \(\left(2\right)\Leftrightarrow7\left(a-b\right)=7\)\(\Rightarrow a=b+1\)

\(\left(1\right)\Leftrightarrow2b+1=7\Leftrightarrow2b=6\Leftrightarrow b=3\)

\(\sqrt{3x^2-2x+2}=3\Rightarrow3x^2-2x+2=9\)

\(\Rightarrow x=\dfrac{2\pm\sqrt{88}}{6}\) (thỏa)

bn nào trả mình VP để mình làm với TT

Sr Phong cách này Trất'ss quá t ko theo dc nên t sẽ ghi full x,y,z nhé chỗ nào cần sửa thành a,b,c thì sửa hộ .-.

Dự đoán khi \(x=y=z=1\) thì \(T_{Max}=1\). Ta chứng minh \(T=1\) là GTLN

\(\dfrac{1}{x^2+y^2+1}\le\dfrac{1}{9}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+1\right)\Rightarrow T\le\dfrac{1}{9}\left(2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)+3\right)\)

Hay cần chứng minh \(\dfrac{1}{9}\left(2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)+3\right)\le1\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\le3=xy+yz+xz\).

\(\Leftrightarrow\left(xy+yz+xz\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)^2\)

WLOG \(x\ge y \ge z\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2}\le\dfrac{1}{y^2}\le\dfrac{1}{z^2}\\xy\ge yz\ge xz\end{matrix}\right.\)

Áp dụng BĐT Chebyshev và AM-GM:

\(VT=\left(xy+yz+xz\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\)

\(\ge3\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)\ge3\sqrt[3]{\dfrac{y}{x}\cdot\dfrac{z}{y}\cdot\dfrac{x}{z}}=9\)

\(\Leftrightarrow9\ge\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)^2\Leftrightarrow3\ge\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\) *đúng*

Hm... mong là vậy :|

\(VT=x^8-x^5+x^2-x+1\)

\(=x^8-2x^4\cdot\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^2}{2}+x^4-x+1\)

\(=\left(x^4-\dfrac{x}{2}\right)^2+\left(\dfrac{x}{2}-1\right)^2+\dfrac{x^2}{2}>0\)

Hay pt vô nghiệm .-.

b)Áp dụng BĐT AM-GM ta có:

\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\sqrt{\dfrac{\sqrt{a}}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}}=2\)

Xảy ra khi \(a=b\)

c)Áp dụng BĐT \(x^2+y^2\ge2xy\) có:

\(VT=\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)

\(\ge2\sqrt{\left(a+b\right)\cdot2\sqrt{ab}}=2\sqrt{2\left(a+b\right)\cdot\sqrt{ab}}=VP\)

Xảy ra khi \(a=b\)