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\(sigma\frac{a}{1+b-a}=sigma\frac{a^2}{a+ab-a^2}\ge\frac{\left(a+b+c\right)^2}{a+b+c+\frac{\left(a+b+c\right)^2}{3}-\frac{\left(a+b+c\right)^2}{3}}=1\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

\(\frac{1}{b^2+c^2}=\frac{1}{1-a^2}=1+\frac{a^2}{b^2+c^2}\le1+\frac{a^2}{2bc}\)

Tương tự cộng lại quy đồng ta có đpcm 

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

làm bừa thôi :) 

Do abc=1 nên ta có thể đặt \(\left(a;b;c\right)=\left(\frac{yz}{x^2};\frac{zx}{y^2};\frac{xy}{z^2}\right)\) ( trong đó \(x^2\ne yz;y^2\ne zx;z^2\ne xy\) ) 

\(VT=sigma\left(\frac{a}{a-1}\right)^2=sigma\left(\frac{yz}{yz-x^2}\right)^2=sigma\left(\frac{x^2}{yz-x^2}+1\right)^2\)

\(\ge\frac{\left(\frac{x^2}{yz-x^2}+\frac{y^2}{zx-y^2}+\frac{z^2}{xy-z^2}+3\right)^2}{3}\ge\frac{9}{3}=3>1\)

\(\left(\frac{a}{a-1};\frac{b}{b-1};\frac{c}{c-1}\right)\rightarrow\left(x;y;z\right)\)

\(\Rightarrow\)\(a=\frac{x}{x-1};b=\frac{y}{y-1};c=\frac{z}{z-1}\)\(\Rightarrow\)\(xyz=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

\(\Leftrightarrow\)\(xy+yz+zx=x+y+z-1\)

\(\Rightarrow\)\(\left(\frac{a}{a-1}\right)^2+\left(\frac{b}{b-1}\right)^2+\left(\frac{c}{c-1}\right)^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\)

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

Dấu "=" xảy ra khi \(abc=\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}=1\) ( quy đồng ra ko biết có đc j ko, bn tự làm nhé ) 

Let \(\left(a;b;c\right)\rightarrow\left(\frac{yz}{x^2};\frac{xz}{y^2};\frac{xy}{z^2}\right)\)  we have:

\(\frac{x^4}{y^2z^2+x^2yz+x^4}+\frac{y^4}{x^2z^2+xy^2z+y^4}+\frac{z^4}{x^2y^2+xyz^2+z^4}\ge1\left(○\right)\)

By Cauchy-Schwarz: \(L-H-S_{\left(○\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{Σ_{cyc}x^4+Σ_{cyc}x^2yz+Σ_{cyc}y^2z^2}\)

Hence we need to prove: \(\frac{\left(x^2+y^2+z^2\right)^2}{Σ_{cyc}x^4+Σ_{cyc}x^2yz+Σ_{cyc}y^2z^2}\ge1\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\geΣ_{cyc}x^4+Σ_{cyc}x^2yz+Σ_{cyc}y^2z^2\)

\(\Leftrightarrow x^2yz+xyz^2+xy^2z\ge x^2y^2+y^2z^2+z^2x^2\)

Follow AM-GM's ineq, it's enough to prove the last ineq

The equality occurs when \(a=b=c=1\)

a/ Đề sai, đề đúng phải là \(p=\frac{a+b+c}{2}\)

b/ \(\Leftrightarrow\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}\ge2\)

\(VT=1-\frac{a^2b}{1+1+a^2b}+1-\frac{b^2c}{1+1+b^2c}+1-\frac{c^2a}{1+1+c^2a}\)

\(VT\ge3-\left(\frac{a^2b}{3\sqrt[3]{a^2b}}+\frac{b^2c}{3\sqrt[3]{b^2c}}+\frac{c^2a}{3\sqrt[3]{c^2a}}\right)\)

\(VT\ge3-\frac{1}{9}\left(3\sqrt[3]{a^2.ab.ab}+3\sqrt[3]{b^2.bc.bc}+3\sqrt[3]{c^2.ca.ca}\right)\)

\(VT\ge3-\frac{1}{9}\left(a^2+2ab+b^2+2bc+c^2+2ca\right)\)

\(VT\ge3-\frac{1}{9}\left(a+b+c\right)^2=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(1-\frac{a^2b}{2+a^2b}\ge1-\frac{a^2b}{3.\sqrt[3]{a^2b}}\)\(\rightarrow1-3\sqrt[3]{a^4b^2}=3.\sqrt[3]{ab.ab.a^2}\rightarrow.....\)

BĐT cần chứng minh tương đương với \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Áp dụng BĐT Cauchy ta có: \(2+a^2b=1+1+a^2b\ge3\sqrt[3]{a^2b}\)

Do đó ta được \(\frac{a^2b}{1+a^2b}\le\frac{a^2b}{3\sqrt[3]{a^2b}}=\frac{a\sqrt[3]{ab^2}}{3}\)

Hoàn toàn tương tự ta được \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le\frac{a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}}{3}\)

Cũng theo BĐT Cauchy ta được \(\sqrt[3]{ab^2}\le\frac{a+b+b}{3}=\frac{a+2b}{3}\)

\(\Rightarrow a\sqrt[3]{ab^2}\le\frac{a\left(a+2b\right)}{3}=\frac{a^2+2ab}{3}\)

Tương tự cũng được \(a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}\le\frac{\left(a+b+c\right)^2}{3}=3\)

Từ đó ta được\(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Vậy BĐT được chứng minh. Dấu "=" xảy ra <=> a=b=c=1

1njfnjgjggnvfkgnbmvfvm 

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow xyz=1\)

Không khó để chứng minh \(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\ge x+y+z\)

\(VT=\Sigma\frac{y^2z}{x^2\left(1+2z\right)}=\Sigma\frac{\left(\frac{y^2}{x^2}\right)}{\frac{1+2z}{z}}\ge\frac{\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)^2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+6}\)

\(\ge\frac{\left(x+y+z\right)^2}{xy+yz+zx+6}\ge\frac{\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}+6}\)

Đặt \(t=x+y+z\ge3\sqrt[3]{xyz}=3\).Cần chứng minh:

\(f\left(t\right)=\frac{t^2}{\frac{t^2}{3}+6}\ge1\Leftrightarrow\frac{2}{3}\left(t-3\right)\left(t+3\right)\ge0\)(đúng)

IS that true?

Làm xong em mới nhận ra không cần đổi biến:D

Ta có:

\(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a^2}{bc}}=3\sqrt[3]{\frac{a^3}{abc}}=3a\)

Tương tự: \(\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\ge3b;\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\ge3c\)

Cộng theo vế 3 BĐT trên suy ra \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c\)

Trở lại bài toán: \(VT=\Sigma_{cyc}\frac{\left(\frac{a^2}{b^2}\right)}{c+2}\ge\frac{\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2}{a+b+c+6}\ge\frac{\left(a+b+c\right)^2}{a+b+c+6}=\frac{t^2}{t+6}\)(với \(t=a+b+c\ge3\sqrt[3]{abc}=3\))

Cần chúng minh: \(\frac{t^2}{t+6}\ge1\Leftrightarrow t^2-t-6\ge0\Leftrightarrow\left(t-3\right)\left(t+2\right)\ge0\)(đúng)