Question

cho a,b,c là số thực dương thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3.\)CMR : \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}+\frac{1}{2}\left(ab+bc+ca\right)\ge3\)

Answer 1

\(\frac{a}{1+b^2}=\frac{a\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự:

\(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)

Cộng lại:

\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{ab}{2}-\frac{bc}{2}-\frac{ca}{2}\)

\(\Rightarrow VT\ge a+b+c\)

Mặt khác:

\(\frac{9}{a+b+c}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\Rightarrow9\le3\left(a+b+c\right)\Rightarrow a+b+c\ge3\)

Khi đó:

\(VT\ge a+b+c\ge3\left(đpcm\right)\)

Dấu "=" xảy ra tại \(a=b=c=1\)

Answer 2

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