Cho $x,y,z>0$, chứng minh rằng:
$\frac{x+y+z}{3\sqrt{3}}\geqslant \frac{xy+yz+zx}{\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}}$
$\frac{x+y+z}{3\sqrt{3}}\geqslant \frac{xy+yz+zx}{\sum \sqrt{x^2+xy+y^2}}$
Started By Minhnguyenthe333, 23-07-2016 - 18:41
#2
Posted 23-07-2016 - 19:27
Math Master
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Cho $x,y,z>0$, chứng minh rằng:
$\frac{x+y+z}{3\sqrt{3}}\geqslant \frac{xy+yz+zx}{\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}}$
Ta luôn có $\sqrt{x^2+xy+y^2} \geq \frac{\sqrt{3}(x+y)}{2}$
Tương Tự : ...
$=> VP \leq \frac{xy+yz+xz}{\sqrt{3}(x+y+z)}$
BĐT quy về CM $\frac{x+y+z}{3} \geq \frac{xy+yz+xz}{x+y+z} <=> (x+y+z)^2 \geq 3(xy+yz+xz)$ (Luôn đúng)
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