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\(=\sqrt{3}\)

Dự đoán khi \(x=y=z=\sqrt{3}\) vậy dc GTNN là \(\frac{3\sqrt{3}}{2}\), cần c/m: \(P\ge\frac{3\sqrt{3}}{2}\)

\(\LeftrightarrowΣ\frac{y^2z^2}{x\left(y^2+z^2\right)}\ge\frac{3}{2}\sqrt{\frac{3}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}}\)

\(\LeftrightarrowΣ\frac{y^3z^3}{y^2+z^2}\ge\frac{3}{2}\sqrt{\frac{3x^4y^4z^4}{x^2y^2+x^2z^2+y^2z^2}}\).Đặt \(\hept{\begin{cases}yz=a\\xz=b\\xy=c\end{cases}}\)

Khi đó ta cần chứng minh \(Σ\frac{a^3}{\frac{ac}{b}+\frac{ab}{c}}\ge\frac{3}{2}\sqrt{\frac{3a^2b^2c^2}{a^2+b^2+c^2}}\)

\(\LeftrightarrowΣ\frac{a^2}{b^2+c^2}\ge\frac{3}{2}\sqrt{\frac{3}{a^2+b^2+c^2}}\) và từ BĐT thuần nhất cuối , ta có thế khẳng định rằng \(a^2+b^2+c^2=3\)

Có nghĩa là ta cần c/m \(Σ\frac{a}{3-a^2}\ge\frac{3}{2}\LeftrightarrowΣ\left(\frac{a}{3-a^2}-\frac{1}{2}\right)\ge0\)

\(\LeftrightarrowΣ\frac{\left(a-1\right)\left(a+3\right)}{3-a^2}\ge0\)\(\LeftrightarrowΣ\left(\frac{\left(a-1\right)\left(a+3\right)}{3-a^2}-\left(a^2-1\right)\right)\ge0\)

\(\LeftrightarrowΣ\frac{a\left(a+2\right)\left(a-1\right)^2}{3-a^2}\ge0\) . XOng!

hùi nãy mem nào k sai cho t T_T t buồn 

\(VT\ge6\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-2\left(xy+yz+zx\right)+2.\frac{9}{4\left(x+y+z\right)}\)

\(=6\left(x+y+z\right)^2-2.\frac{\left(x+y+z\right)^2}{3}+\frac{9}{2\left(x+y+z\right)}=6.\left(\frac{3}{4}\right)^2-2.\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{9}{2.\frac{3}{4}}\)

\(=\frac{27}{8}-\frac{3}{8}+6=9\)

\(\Rightarrow\)\(VT\ge9\) ( đpcm ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)

Chúc bạn học tốt ~