kho ghe

\(a+b+c=1\)

\(P=\frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2}\)

\(P=\frac{y^2z^2}{x\left(y^2+z^2\right)}+\frac{z^2x^2}{y\left(x^2+z^2\right)}+\frac{x^2y^2}{z\left(x^2+y^2\right)}\)

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

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\) thì \(a^2+b^2+c^2=1\) Ta cần chứng minh:

\(P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)

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

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

Theo đánh giá bởi AM - GM ta có:

\(a^2\left(1-a^2\right)^2=\frac{1}{2}\cdot2a^2\cdot\left(1-a^2\right)\left(1-a^2\right)\)

\(\le\frac{1}{2}\left(\frac{2a^2+1-a^2+1-a^2}{3}\right)^3=\frac{4}{27}\)

\(\Rightarrow a\left(1-a^2\right)^2\le\frac{2}{3\sqrt{3}}\Leftrightarrow\frac{a^2}{a\left(1-a\right)^2}\ge\frac{3\sqrt{3}}{2}a^2\)

Tương tự rồi cộng lại ta có ngay điều phải chứng minh

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

Ta có \(a,b,c>0;a^2+b^2+c^2=1\)

và \(P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)

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

Áp dụng bất đẳng thức Cô-si cho 3 số dương ta có

\(a^2\left(1-a^2\right)^2=\frac{1}{2}.2a^2.\left(1-a^2\right)\left(1-a^2\right)\)

\(\le\frac{1}{2}\left(\frac{2a^2+1-a^2+1-a^2}{3}\right)^3=\frac{4}{27}\)

\(\Rightarrow a\left(1-a^2\right)\le\frac{2}{3\sqrt{3}}\Rightarrow\frac{a^2}{a\left(1-a^2\right)}\ge\frac{3\sqrt{3}}{2}a^2\)(1)

Tương tự \(\frac{b^2}{b\left(1-b^2\right)}\ge\frac{3\sqrt{3}}{2}b^2\)(2)

\(\frac{c^2}{c\left(1-c^2\right)}\ge\frac{3\sqrt{3}}{2}c^2\)(3)

từ (1),(2) và (3) ta có \(P\ge\frac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\frac{3\sqrt{3}}{2}\)

Vậy Min của \(P=\frac{3\sqrt{3}}{2}\)Khi x=y=z\(=\sqrt{3}\)

Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)