đề toán chuyên Lam Sơn

Question

cho $x,y,z>0; x+y+z\leq \frac{3}{2}$. tìm gtnn của: $P=\frac{x(yz+1)^2}{z^2(xz+1)}+\frac{y(xz+1)^2}{x^2(xy+1)}+\frac{z(xy+1)^2}{y^2(yz+1)}$

❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ · Answer 1 · 6/11/2016 1:12:31 PM

Ta có: $P = \sum \frac{x (y z + 1)^{2}}{z^{2} (x z + 1)}$

$= \sum \frac{\frac{{(y z + 1)}^{2}}{z^{2}}}{\frac{(x z + 1)}{x}}$

$= \sum \frac{{(y + \frac{1}{z})}^{2}}{z + \frac{1}{x}}$

$\geq \frac{{(x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})}^{2}}{x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$

$= x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq x + y + z + \frac{9}{x + y + z}$

$= (x + y + z + \frac{9}{4 (x + y + z)}) + \frac{27}{4 (x + y + z)}$

$\geq 2 \sqrt{\frac{9}{4}} + \frac{27}{4. \frac{3}{2}} = \frac{15}{2}$

Trả lời 11-06-16 01:12 PM

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