Lightning Farron

djnh lam` nhung thay lop 8 nen thoi so mn ko hieu ;(

Đặt \(a^{\dfrac{1}{9}};b^{\dfrac{1}{9}};c^{\dfrac{1}{9}}\rightarrow x;y;z\)\(\left(x;y;z>0;xyz=1\right)\)

Ta có BĐT:\(\dfrac{1}{\sqrt{8x^9+1}}\ge\dfrac{1}{x^8+x^4+1}\)

\(\Leftrightarrow\dfrac{\dfrac{\left(x-1\right)^2x^4\left(x^{10}+2x^9+3x^8+4x^7+7x^6+10x^5+13x^4+8x^3+6x^2+4x+2\right)}{\left(x^2-x+1\right)^2\left(x^2+x+1\right)^2\left(2x^3+1\right)\left(x^4-x^2+1\right)^2\left(4x^6-2x^3+1\right)}}{\dfrac{1}{\sqrt{8x^9+1}}+\dfrac{1}{x^8+x^4+1}}\ge0\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(A\ge\dfrac{1}{x^8+x^4+1}+\dfrac{1}{y^8+y^4+1}+\dfrac{1}{z^8+z^4+1}\ge1\)

Dấu "=" khi \(x=y=z=a=b=c=1\)

Ôi vãi thức khuya đá banh luôn .-. ghê zz

Cho \(x;y;z\) không âm thỏa mãn \(x+y^2+z^3=1\). Tìm GTLN của \(x^2y+y^2z+z^2x\)

\(\bbox[5px,border:2px solid #C0A000]{\text{Top 1: 0 điểm }} \)

\(\dfrac{1}{a^2+b^2+2}+\dfrac{1}{b^2+c^2+2}+\dfrac{1}{c^2+a^2+2}\le\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{a^2+b^2}{a^2+b^2+2}+\dfrac{b^2+c^2}{b^2+c^2+2}+\dfrac{c^2+a^2}{c^2+a^2+2}\ge\dfrac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\)

\(\ge\dfrac{\sqrt{3\left(a^2b^2+b^2c^2+a^2c^2\right)}+2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\)

\(\ge\dfrac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\)

Cần chứng minh \(\dfrac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge0\) *luôn đúng*

Sửa: \(a;b>0\)

Áp dụng BĐT AM-GM ta có:

\(A=\left(a+1\right)\left(1+\dfrac{1}{b}\right)+\left(b+1\right)\left(1+\dfrac{1}{a}\right)\)

\(=\dfrac{a}{b}+\dfrac{b}{a}+a+\dfrac{1}{a}+b+\dfrac{1}{b}+2\)

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

\(\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}+2\sqrt{a\cdot\dfrac{1}{2a}}+2\sqrt{b\cdot\dfrac{1}{2b}}+2\sqrt{\dfrac{1}{2a}\cdot\dfrac{1}{2b}}+2\)

\(=4+2\sqrt{2}+\dfrac{1}{\sqrt{ab}}\)\(\ge4+2\sqrt{2}+\dfrac{1}{\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)

\(=4+3\sqrt{2}\)

Dấu \("="\) xảy ra khi \(a=b=\dfrac{1}{\sqrt{2}}\)

Áp dụng BĐT AM-GM ta có:

\(P=x+y+\dfrac{6}{x}+\dfrac{24}{y}\)

\(=x+\dfrac{4}{x}+y+\dfrac{16}{y}+\dfrac{2}{x}+\dfrac{8}{y}\)

\(\ge2\sqrt{x\cdot\dfrac{4}{x}}+2\sqrt{y\cdot\dfrac{16}{y}}+2\left(\dfrac{1}{x}+\dfrac{4}{y}\right)\)

\(\ge4+8+2\dfrac{\left(1+2\right)^2}{x+y}\ge15\)

Xảy ra khi \(x=2;y=4\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\dfrac{1}{\sqrt{x}+2\sqrt{y}}\le\dfrac{1}{9}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{y}}\right)\)

Tương tự cho 2 BĐT trên ta có:

\(\dfrac{1}{3}VP\le\dfrac{1}{9}\cdot3\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)=\dfrac{1}{3}VT\)

Xảy ra khi \(x=y=z\)

Giả sử \(x=a;y=b;z=c\)

Ta có: \(\dfrac{2x}{a}+\dfrac{3y}{b}+\dfrac{4z}{c}\ge9\sqrt[9]{\dfrac{x^2y^3z^4}{a^2b^3c^4}}\)

Mà \(\left(\dfrac{2x}{a}+\dfrac{3y}{b}+\dfrac{4z}{c}\right)^2\le\left(x^2+y^2+z^2\right)\left(\dfrac{4}{a^2}+\dfrac{9}{b^2}+\dfrac{16}{c^2}\right)\)

Xảy ra khi \(\dfrac{ax}{2}=\dfrac{by}{3}=\dfrac{cz}{4}\Leftrightarrow\dfrac{a^2}{2}=\dfrac{b^2}{3}=\dfrac{c^2}{4}\)

Ta có hệ \(\left\{{}\begin{matrix}\dfrac{a^2}{2}=\dfrac{b^2}{3}=\dfrac{c^2}{4}\\a^2+b^2+c^2=1\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{\sqrt{2}}{3};b=\dfrac{\sqrt{3}}{3};c=\dfrac{2}{3}\)

Vậy \(Max_P=\dfrac{32\sqrt{3}}{6561}\) khi \(x=\dfrac{\sqrt{2}}{3};y=\dfrac{\sqrt{3}}{3};z=\dfrac{2}{3}\)

Từ \(pt\left(2\right)\Leftrightarrow\left(2x+4y-1\right)^2\left(2x-y-1\right)=\left(4x-2y-3\right)^2\left(x+2y\right)\)

\(\Leftrightarrow-\left(x-3y-1\right)\left(8x^2-8y^2-4x-8y+12xy-1\right)=0\)

tự làm nốt đi (nóng quááááááááááááááá)

Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c>0\end{matrix}\right.\)

Và \(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}+\dfrac{bc}{\sqrt{b^2+c^2+2a^2}}+\dfrac{ca}{\sqrt{c^2+a^2+2b^2}}\le\dfrac{1}{2}\)

Ta có:\(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}=\dfrac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)

\(\le\dfrac{2ab}{a+b+2c}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)\)

\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)

Dấu "=" khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=\dfrac{1}{9}\)

Ta có:

\(\dfrac{1}{ab+a+2}\le\dfrac{1}{4}\left(\dfrac{1}{ab+1}+\dfrac{1}{a+1}\right)=\dfrac{1}{4}\left(\dfrac{c}{1+c}+\dfrac{1}{a+1}\right)\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\le\dfrac{1}{4}\left(\dfrac{a+1}{a+1}+\dfrac{b+1}{b+1}+\dfrac{c+1}{c+1}\right)=\dfrac{3}{4}\)

Add: \(x;y;z>0\)

G.sử dấu "=" xảy ra khi \(x=y=a;z=b\)

Áp dụng BĐT AM-GM ta có:

\(abx^2+aby^2\ge2abxy\)

\(b^2y^2+a^2z^2\ge2abyz\)

\(b^2x^2+a^2z^2\ge2abxz\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(ab+b^2\right)+2a^2z^2\ge2ab\left(xy+yz+xz\right)=2ab\)

Ta có hệ \(\left\{{}\begin{matrix}ab+b^2=6a^2\\a^2+2ab=1\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{1}{\sqrt{5}};b=\dfrac{2}{\sqrt{5}}\)

Vậy \(A_{Min}=2\) khi \(x=y=\dfrac{1}{\sqrt{5}};z=\dfrac{2}{\sqrt{5}}\)

Ta có:\(\left(9x^3+3y^2+z\right)\left(\dfrac{1}{9x}+\dfrac{1}{3}+z\right)\ge\left(x+y+z\right)^2\)

\(\Leftrightarrow\dfrac{x}{9x^3+3y^2+z}\le\dfrac{x\left(\dfrac{1}{9x}+\dfrac{1}{3}+z\right)}{\left(x+y+z\right)^2}=\dfrac{\dfrac{1}{9}+\dfrac{x}{3}+xz}{\left(x+y+z\right)^2}\)

Tương tự rồi cộng theo vế:

\(Σ_{cyc}\dfrac{x}{9x^3+3y^2+z}\le\dfrac{\dfrac{1}{9}\cdot3+\dfrac{x+y+z}{3}+xy+yz+xz}{\left(x+y+z\right)^2}\)

\(\le\dfrac{\dfrac{1}{9}\cdot3+\dfrac{x+y+z}{3}+\dfrac{\left(x+y+z\right)^2}{3}}{\left(x+y+z\right)^2}=1\)

Lại có: \(2017\left(xy+yz+xz\right)\le2017\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{2017}{3}\)

\(\Rightarrow A\le\dfrac{2020}{3}\)

Dấu "=" khi \(x=y=z=\dfrac{1}{3}\)

Vậy ko ra yếu zzzz

Cho \(a;b;c>0\). CMR \(\dfrac{a}{\sqrt{a^2+3bc}}+\dfrac{b}{\sqrt{b^2+3ac}}+\dfrac{c}{\sqrt{c^2+3ab}}\le\dfrac{9(a^2+b^2+c^2)}{2(a+b+c)^2}\)

Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)

Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)

\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)

\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)

Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)

\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)

\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*

Cho \(a,b,c\) dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=8\). Chứng minh

\(\dfrac{a+b+c}{3}\ge\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}\)

eztosol

Đặt \(a=\dfrac{kx}{y};b=\dfrac{ky}{z};c=\dfrac{kz}{x}\Rightarrow abc=k^3\)

Ta có: \(BDT\Leftrightarrow\dfrac{yz}{kx\left(ky+z\right)}+\dfrac{xz}{ky\left(kz+x\right)}+\dfrac{xy}{kz\left(kx+y\right)}\ge\dfrac{3}{1+k^3}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT=\dfrac{y^2z^2}{kxyz\left(ky+z\right)}+\dfrac{x^2z^2}{kxyz\left(kz+x\right)}+\dfrac{x^2y^2}{kxyz\left(kx+y\right)}\)

\(\ge\dfrac{\left(xy+yz+xz\right)^2}{xyz\left(x+y+z\right)k\left(k+1\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{xyz\left(x+y+z\right)k\left(k+1\right)}=\dfrac{3}{k\left(k+1\right)}\)

Cần chứng minh \(\dfrac{3}{k\left(k+1\right)}\ge\dfrac{3}{1+k^3}\)

\(\Leftrightarrow\dfrac{3\left(k-1\right)^2}{k\left(k+1\right)\left(k^2-k+1\right)}\ge0\) (luôn đúng)

đặt x/y=a hay xy/z=a hay j đó là ra nói chung là 4 biế
n lười nháp

Áp dụng BĐT AM-GM ta có:

\(\dfrac{abc}{a^2+bc}\le\dfrac{abc}{2a\sqrt{bc}}=\dfrac{\sqrt{bc}}{2}\le\dfrac{b+c}{4}\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(abc.VT\le\dfrac{2\left(a+b+c\right)}{4}=1\Leftrightarrow VT\le\dfrac{1}{abc}=VP\)

Dấu "="\(\Leftrightarrow a=b=c=\dfrac{2}{3}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\dfrac{ab}{\sqrt{3c+ab}}=\dfrac{ab}{\sqrt{\left(a+b+c\right)c+ab}}=\dfrac{ab}{\sqrt{c^2+ab+bc+ca}}\)

\(=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(P\le\dfrac{1}{2}\left(a+b+c\right)=\dfrac{3}{2}\)

\("="\Leftrightarrow a=b=c=1\)

\(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\)

\(\ge2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}+1=3\)

Khi x=1

Sửa \(y+\sqrt{2015+x^2}\rightarrow y+\sqrt{2015+y^2}\)

Ta có: \(\left(x+\sqrt{2015+x^2}\right)\left(y+\sqrt{2015+y^2}\right)=2015\)

\(\Leftrightarrow\left(x+\sqrt{2015+x^2}\right)\left(\sqrt{2015+x^2}-x\right)\left(y+\sqrt{2015+y^2}\right)=2015\left(\sqrt{2015+x^2}-x\right)\)

\(\Leftrightarrow2015\left(y+\sqrt{2015+y^2}\right)=2015\left(\sqrt{2015+x^2}-x\right)\)

\(\Leftrightarrow x+y=\sqrt{2015+x^2}-\sqrt{2015+y^2}\)

Tương tự ta cũng có: \(x+y=\sqrt{2015+y^2}-\sqrt{2015+x^2}\)

Cộng theo vế 2 đẳng thức trên ta có:

\(2\left(x+y\right)=0\Leftrightarrow x=-y\)

Thay \(x=-y\) vào \(pt\left(2\right)\) ta có:

\(23y^2=23\Leftrightarrow y=\pm1\Leftrightarrow x=\mp1\)