Question

[nguyễn minh]

cho a,b,c>0 thỏa mãn a+b+c=3. CMR: \(\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1\)

Answer 1

[Akai Haruma]

Lời giải:

Ta thấy:

\(\text{VT}=\frac{c^2}{2ab^2c^2+c^2}+\frac{a^2}{2bc^2a^2+a^2}+\frac{b^2}{2ca^2b^2+b^2}\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}(2ab^2c^2+c^2+2bc^2a^2+a^2+2ca^2b^2+b^2)\geq (c+a+b)^2\)

\(\Leftrightarrow \text{VT}\geq \frac{(a+b+c)^2}{2abc(ab+bc+ac)+a^2+b^2+c^2}(*)\)

Áp dụng BĐT Am-GM:

\(3=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq 1\)

\(\Rightarrow 2abc(ab+bc+ac)\leq 2(ab+bc+ac)\)

\(\Rightarrow \frac{(a+b+c)^2}{2abc(ab+bc+ac)+a^2+b^2+c^2}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)+a^2+b^2+c^2}=1(**)\)

Từ \((*); (**)\Rightarrow \text{VT}\geq 1\)

Ta có đpcm. Dấu "=" xảy ra khi $a=b=c=1$

Answer 2

[Akai Haruma]

Cách khác bằng AM-GM:

\(\text{VT}=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)(1)\)

Áp dụng BĐT AM-GM:

\(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}=\frac{2ab^2}{ab^2+ab^2+1}+\frac{2bc^2}{bc^2+bc^2+1}+\frac{2ca^2}{ca^2+ca^2+1}\)

\(\leq \frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^2a^4}}=\frac{2}{3}(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2})\)

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

Từ \((1);(2)\Rightarrow \text{VT}\geq 3-2=1\) (đpcm)