Question

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Cho a,b,c>0 : a²+b²+c²=3

CMR: (a²+1)/b++(b²+1)/c+(c²+1)/a >hoặc = 2(a+b+c)

Answer 1

[Vũ Tiến Manh]

Thay 1=\(\frac{a^2+b^2+c^2}{3}\)vào va rút gọn ta được

VT= \(\frac{4}{3}\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{1}{3}\left(\frac{c^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}\right)+\frac{1}{3}\left(a+b+c\right)\)(1)

Áp dụng \(\frac{x^2}{m}+\frac{y^2}{n}+\frac{z^2}{p}\ge\frac{\left(x+y+z\right)^2}{m+n+p}\left(bunhiacopxky\right)\) ta được

(1) \(\ge\frac{4}{3}\frac{\left(a+b+c\right)^2}{a+b+c}+\frac{1}{3}\frac{\left(a+b+c\right)^2}{a+b+c}+\frac{1}{3}\left(a+b+c\right)=2\left(a+b+c\right).\)

Dấu'=' khi a=b=c