Question

dạ mọi người giúp em bài Toán 8 này với ạ! Dạ em cảm ơn ạ

Cho a,b,c > 0 thỏa mãn abc = 1. Chứng minh

a) \(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}\ge\:3\)

b) \(\frac{1}{a+b^4+c^4}+\frac{1}{b+a^4+c^4}+\frac{1}{c+b^4+a^4}\le\:1\)

Answer 1

a/

\(VT\ge\frac{\frac{1}{2}\left(a+b\right)^2}{a+b}+\frac{\frac{1}{2}\left(b+c\right)^2}{b+c}+\frac{\frac{1}{2}\left(c+a\right)^2}{c+a}=a+b+c\ge3\sqrt[3]{abc}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

b/ Ta có: \(x^4+y^4\ge\frac{1}{2}\left(x^2+y^2\right)\left(y^2+y^2\right)\ge xy\left(x^2+y^2\right)\)

\(\Rightarrow VT\le\frac{1}{a+bc\left(b^2+c^2\right)}+\frac{1}{b+ca\left(a^2+c^2\right)}+\frac{1}{c+ab\left(a^2+b^2\right)}\)

\(VT\le\frac{1}{a+\frac{1}{a}\left(b^2+c^2\right)}+\frac{1}{b+\frac{1}{b}\left(a^2+c^2\right)}+\frac{1}{c+\frac{1}{c}\left(a^2+b^2\right)}\)

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

\(VT\le\frac{a+b+c}{\frac{1}{3}\left(a+b+c\right)^2}=\frac{3}{a+b+c}\le\frac{3}{3\sqrt[3]{abc}}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)