Question

cho sac số thực dương a,b,c thỏa mãn \(a^2+b^2+c^2=3\). chứng minh \(\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{ca^2}{c+a^2}\ge a+b+c\)

Answer 1

Lời giải:

Áp dụng BĐT AM-GM ta có:
\(\text{VT}=2\left(a-\frac{ab^2}{a+b^2}\right)+2\left(b-\frac{bc^2}{b+c^2}\right)+2\left(c-\frac{ca^2}{c+a^2}\right)\)

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

\(\geq 2(a+b+c)-2.\left(\frac{ab^2}{2\sqrt{ab^2}}+\frac{bc^2}{2\sqrt{bc^2}}+\frac{ca^2}{2\sqrt{ca^2}}\right)\)

\(=2(a+b+c)-(\sqrt{ab^2}+\sqrt{bc^2}+\sqrt{ca^2})\)

\(\geq 2(a+b+c)-\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)=a+b+c\)

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c=1$