Lời giải:
$a^2+1=a^2+ab+bc+ac=(a+b)(a+c)$
$b^2+1=b^2+ab+bc+ac=(b+c)(b+a)$
$c^2+1=c^2+ab+bc+ac=(c+a)(c+b)$
Do đó:
\(\text{VT}=\frac{2a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\\ \leq \frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{4(b+c)}+\frac{b}{b+a}+\frac{c}{c+a}+\frac{c}{4(c+b)}\)
\(=\frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{1}{4}.\frac{b+c}{b+c}=1+1+\frac{1}{4}=\frac{9}{4}\)
Vậy ta có đpcm.
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