Lightning Farron

B1: Từ \(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}\)

\(\Leftrightarrow\dfrac{\left(xy-1\right)\left(x-y\right)^2}{\left(x^2+1\right)\left(y^2+1\right)\left(xy+1\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\xy=1\end{matrix}\right.\)

*)Với \(x=y\) khi đó ta có: \(P=\dfrac{1}{x^2+1}+\dfrac{1}{1+x^2}+\dfrac{2}{1+x^2}=\dfrac{4}{x^2+1}\)

*)Với \(xy=1\) khi đó ta có: \(P=\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}+\dfrac{2}{1+1}\)

\(=\dfrac{x+y}{xy\left(x+y\right)}+1=2\)

vẽ cái hình xem nào

\(x;y;z\ge0;p;q;n\in N^{star};\sum_{cyc}x^n=3;2p+2q>2n\text{.Prove}\sum_{cyc}\dfrac{x^p}{y^q}\ge3\)

\(\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}+\dfrac{\left(c+a\right)^2}{5b^2+\left(c+a\right)^2}+\dfrac{\left(a+b\right)^2}{5c^2+\left(a+b\right)}\ge\dfrac{4}{3}\)

\(\Leftrightarrow\dfrac{-20a^2+10bc+5b^2+c^2}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{-20b^2+10ac+5c^2+5a^2}{9\left(5b^2+\left(c+a\right)^2\right)}+\dfrac{-20c^2+10ab+5a^2+5b^2}{9\left(5c^2+\left(a+b\right)\right)}\ge0\)

\(\Leftrightarrow\sum_{cyc}\dfrac{\left(c-a\right)\left(10a+5b+5c\right)-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\dfrac{-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{\left(a-b\right)\left(10b+5a+5c\right)}{9\left(5b^2+\left(a+c\right)^2\right)}\right)\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)\left(\dfrac{10b+5a+5c}{9\left(5b^2+\left(a+c\right)^2\right)}-\dfrac{10a+5b+5c}{9\left(5a^2+\left(b+c\right)^2\right)}\right)\right)\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)^2\dfrac{5\left(a^2+b^2-c^2+4ab\right)}{3\left(a^2+2ac+5b^2+c^2\right)\left(5a^2+b^2+2bc+c^2\right)}\right)\ge0\)

Dau "=" khi \(a=b=c\)

Thì admin cũng cố gắng lắm rồi, còn bận trăm công nghìn việc cố được đến thế nào thì cố còn phải chịu thôi =)) không có thời gian thì làm sao mà duy trì được web tốt được ai vẫn muốn onl thì cứ onl =))

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

Xét \(pt\left(1\right)\Leftrightarrow2x^2+y^2-3xy-4x+3y+2=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(2x-y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=2x-2\end{matrix}\right.\)

*)\(y=x-1\) thay vao \(pt(2)\) :

\(pt\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=0\end{matrix}\right.\)

*)\(y=2x-2\) thay vao \(pt(2)\):

\(pt\Leftrightarrow\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)

\(\Leftrightarrow\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{1}{\sqrt{x-1}}\right)=0\)

\(\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)

Ap dung BDT Cauchy-Schwarz ta co:

\(\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\left(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\right)\ge\left(a^2+b^2+c^2\right)^2\)

Can chung minh \(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\ge a^2+b^2+c^2\)

\(VT=\dfrac{a^4}{ac}+\dfrac{b^4}{ab}+\dfrac{c^4}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)

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

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

DK: \(x\ge5\)

\(BPT\Leftrightarrow x^2-7x+2-2\sqrt{x^2-7x+10}< 0\)

\(\Leftrightarrow t^2-8-2t< 0\)\(\left(t=\sqrt{x^2-7x+10}\ge0\right)\)

\(\Leftrightarrow\left(t+2\right)\left(t-4\right)< 0\)

\(\Leftrightarrow-2< t< 4\)\(\Leftrightarrow-2< \sqrt{x^2-7x+10}< 4\)

\(\Leftrightarrow\sqrt{x^2-7x+10}< 4\)\(\Leftrightarrow x^2-7x-6< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}5\le x< \dfrac{7+\sqrt{73}}{2}\\\dfrac{7-\sqrt{73}}{2}< x\le2\end{matrix}\right.\)

\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)

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

\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)

\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)

\(VT=\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\)

\(\ge\dfrac{4}{x^2+2xy+y^2}\)

\(=\dfrac{4}{\left(x+y\right)^2}>4\)

\(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)

\(\Leftrightarrow A^3=18+3A\)\(\Leftrightarrow A^3-3A-18=0\)

\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)

De thay: \(A^2+3A+6=\left(A+\dfrac{3}{2}\right)^2+\dfrac{15}{4}>0\forall A\)

\(\Leftrightarrow A=3\)

Dễ thấy: \(a,b,c\) là 3 nghiệm của pt \(f\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)=x^3-6x^2+9x-abc\)

Theo định lí Rolle thì \(f'\) có 1 nghiệm trên khoảng \(\left(a,b\right)\) và \(\left(b,c\right)\)

Mà \(f'\left(x\right)=3x^2-12x+9=3\left(x-1\right)\left(x-3\right)\) có 2 nghiệm \(x_1=1;x_2=3\).Do đó:

\(a< x_1=1< b< x_2=3< c\)

Vậy cần cm \(c<4\) . Mặt khác \(c\) đạt GTLN khi \(abc\) đạt GTLN

Xảy ra vì \(a\rightarrow b\rightarrow1\) ta có \(c<4\)

Search 10 bài 10 bài thấy trên mạng why so ez :)

\(BDT\Leftrightarrow2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge5a^2b^2c+5a^2bc^2+5ab^2c^2\)

Ta chứng minh được \(ab^4+bc^4+ca^4\ge a^2b^2c+a^2bc^2+ab^2c^2\)

\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

\(VT=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\)

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

Vậy ta cần chứng minh \(2a^4b+2b^4c+2c^4a+2ab^4+2bc^4+2ca^4\ge4a^2b^2c+4a^2bc^2+4ab^2c^2\)

\(\Leftrightarrow\sum_{cyc}\left(2c^3+bc^2-b^2c+ac^2-a^2c+3ab^2+3a^2b\right)\left(a-b\right)^2\ge0\)

Dấu "=" xảy ra khi \(a=b=c\)

SOS helps ^^

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

\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}-b+\dfrac{b^2+c^2}{a+c}-c+\dfrac{c^2+a^2}{a+b}-a\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a+c\right)-\left(c-a\right)\left(a+b\right)}{b+c}\ge0\)

\(\Leftrightarrow\sum\left(a-b\right)\left(\dfrac{a+c}{b+c}-\dfrac{b+c}{a+c}\right)\ge0\)

\(\Leftrightarrow\sum\left(a-b\right)^2\dfrac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge0\)

Cần thêm \(a;b;c\) dương nha

sos helps :3

chứng minh \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2\)

cuối cùng cũng đến ngày này T.T

woa hai người qua mấy tỉ năm vẫn thù nhau à :)

cuoc thi ngu van ma nguoi to chuc go qua nhieu loi chinh ta :D nhin la chan :sosad:

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 \(x=y=z=a=b=c=1\)