Đặt \(P=\frac{a^4}{\left(a+2\right)\left(b+2\right)}+\frac{b^4}{\left(b+2\right)\left(c+2\right)}+\frac{c^4}{\left(c+2\right)\left(a+2\right)}\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a^4}{\left(a+2\right)\left(b+2\right)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\ge4\sqrt[4]{\frac{a^2}{\left(a+2\right)\left(b+2\right)}.\frac{a+2}{27}.\frac{b+2}{27}.\frac{1}{9}}=\frac{4a}{9}\)(1)

\(\frac{b^4}{\left(b+2\right)\left(c+2\right)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\ge4\sqrt[4]{\frac{b^2}{\left(b+2\right)\left(c+2\right)}.\frac{b+2}{27}.\frac{c+2}{27}.\frac{1}{9}}=\frac{4b}{9}\)(2)

\(\frac{c^4}{\left(c+2\right)\left(a+2\right)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\ge4\sqrt[4]{\frac{c^2}{\left(c+2\right)\left(a+2\right)}.\frac{c+2}{27}.\frac{a+2}{27}.\frac{1}{9}}=\frac{4c}{9}\)(3)

Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\)ta được:

\(P+\frac{2\left(a+b+c\right)+12}{27}+\frac{3}{9}\ge\frac{4\left(a+b+c\right)}{9}\)

\(\Leftrightarrow P+\frac{2}{3}+\frac{3}{9}\ge\frac{4}{3}\)

\(\Leftrightarrow P\ge\frac{1}{3}\left(đpcm\right)\)Dấu"="xảy ra \(\Leftrightarrow a=b=c=1\)

Cách khác

Ta co:

\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{\Sigma_{cyc}\left(a+2\right)\left(b+2\right)+12}\ge\frac{\left(a+b+c\right)^4}{36\left(a+b+c\right)+9\left(ab+bc+ca\right)+108}\ge\frac{3^4}{108.2+9.\frac{\left(a+b+c\right)^2}{3}}=\frac{1}{3}\)

phương trình đâu vậy?

Phương trình j vậy

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

\(\Leftrightarrow\left(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}\right)^2=4\)

\(\Leftrightarrow x-\sqrt{x^2-1}+2\sqrt{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}+x+\sqrt{x^2-1}=4\)

\(\Leftrightarrow2x+2\sqrt{x^2-x^2+1}=4\)

\(\Leftrightarrow2x+2=4\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

vậy x=1

 Dễ ẹt

Ghi theo tớ là đc 10+ đấy:

Trả lời:Em quên mang phao mong cô thông cảm

\(P\le\Sigma_{cyc}\frac{1}{ab\left(a+b\right)+abc}=\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{abc}\)