Question

[Phú Nguyễn Đinh]

cho a,b,c là 3 số thực dương. chứng minh rằng : \(\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\dfrac{b^3}{b^3+\left(c+a\right)^3}}+\sqrt{\dfrac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)

Answer 1

[missing you =]

\(\Sigma\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}=\Sigma\sqrt{\dfrac{1}{1+\left(\dfrac{b+c}{a}\right)^3}}\)\(\left(1\right)\)

\(đặt:\left(\left(\dfrac{b+c}{a}\right)^{ };\left(\dfrac{c+a}{b}\right)^{ };\left(\dfrac{a+b}{c}\right)^{ }\right)=\left(x;y;z\right)\)

\(\left(1\right)\Leftrightarrow\sqrt{\dfrac{1}{1+x^3}}+\sqrt{\dfrac{1}{1+y^3}}+\sqrt{\dfrac{1}{1+z^3}}=\sqrt{\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}}+\sqrt{\dfrac{1}{\left(y+1\right)\left(y^2-y+1\right)}}+\sqrt{\left(z+1\right)\left(z^2-z+1\right)}\)

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

\(tương\) \(tự\Rightarrow\left(1\right)\ge\dfrac{2}{x^2+2}+\dfrac{2}{y^2+2}+\dfrac{2}{z^2+2}\)

\(=\dfrac{2}{\left(\dfrac{b+c}{a}\right)^2+2}+\dfrac{2}{\left(\dfrac{c+a}{b}\right)^2+2}+\dfrac{2}{\left(\dfrac{a+b}{c}\right)^2+2}=\dfrac{2a^2}{\left(b+c\right)^2+2a^2}+\dfrac{2b^2}{\left(c+a\right)^2+2b^2}+\dfrac{2c^2}{\left(a+b\right)^2+2c^2}\)

\(bunhia\Rightarrow\left(b+c\right)^2\le2\left(b^2+c^2\right)\Rightarrow\dfrac{2a^2}{\left(b+c\right)^2+2a^2}\ge\dfrac{2a^2}{2\left(a^2+b^2\right)+2a^2}=\dfrac{a^2}{a^2+b^2+c^2}\)

\(tương\) \(tự\Rightarrow\left(1\right)\ge\dfrac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\left(đpcm\right)\)