Ta có $(x^2+y^2+z^2)(x+y+z)$=$x^2y+y^2z+z^2x$
+$\sum_{cyc}^{cyc}(x^3+xy^2)\geq 3(x^2y+y^2z+z^2x)$(với$ \sum_{cyc}^{cyc}(x^3+xy^2) \geq 2(x^2y+y^2z+z^2x) $( áp dụng BĐT Cau chy cho từng cặp )
$=> x^2y+y^2z+z^2x \leq x^2+y^2+z^2$
Lại có :$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx=9$
$=> xy+yz+zx=\frac{9-x^2+y^2+z^2}{2}$
$=> P \geq (x^2+y^2+z^2) +\frac{9}{2(x^2+y^2+z^2)}-\frac{1}{2}$
$P\geq t+\frac{9}{2t} -\frac{1}{2}$ ( với$ t=x^2+y^2+z^2$)
$P \geq \frac{t}{2}+\frac{9}{2t}+\frac{t}{2}-\frac{1}{2}$
Áp dụng BĐT AM-GM ta có:
$P \geq 3+\frac{t}{2}-\frac{1}{2}=4$
dấu "=" xảy ra khi $x=y=z=1$
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