ta có $P=\frac{(x+y)^{2}}{z^{2}+4(xy+yz+zx)}=\frac{(x+y)^{2}}{z^{2}+4z(x+y)+(x+y)^{2}}$ $=\frac{(\frac{x}{z}+\frac{y}{z})^{2}}{1+4(\frac{x}{z}+\frac{y}{z})+(\frac{x}{z}+\frac{y}{z})^{2}}$Đặt $t=\frac{x+y}{z} (t\in \left[ 1{;}4 \right])$ khi đó P$=\frac{t^{2}}{1+4t+t^{2}}$ ta cần cm $P\geq \frac{1}{6}$ $\Leftrightarrow 5t^{2}-4t-1\geq0 \Leftrightarrow (t-1)(5t+1)\geq0$ ( lđ vs $\forall t\in \left[ 1{;}4 \right]$) Dấu "=" $\Leftrightarrow x=y=1;z=2$
ta có $P=\frac{(x+y)^{2}}{z^{2}+4(xy+yz+zx)}=\frac{(x+y)^{2}}{z^{2}+4z(x+y)+(x+y)^{2}}$ $=\frac{(\frac{x}{z}+\frac{y}{z})^{2}}{1+4(\frac{x}{z}+\frac{y}{z})+(\frac{x}{z}+\frac{y}{z})^{2}}$Đặt $t=\frac{x+y}{z} (t\in \left[ 1{;}4 \right])$ khi đó P$=\frac{t^{2}}{1+4t+t^{2}}$ ta cần cm $P\geq \frac{1}{6}$ $\Leftrightarrow 5t^{2}-4t-1\geq0 \Leftrightarrow (t-1)(5t+1)\geq0$ ( lđ vs $\forall t\in \left[ 1{;}4 \right]$)
ta có $P=\frac{(x+y)^{2}}{z^{2}+4(xy+yz+zx)}=\frac{(x+y)^{2}}{z^{2}+4z(x+y)+(x+y)^{2}}$ $=\frac{(\frac{x}{z}+\frac{y}{z})^{2}}{1+4(\frac{x}{z}+\frac{y}{z})+(\frac{x}{z}+\frac{y}{z})^{2}}$Đặt $t=\frac{x+y}{z} (t\in \left[ 1{;}4 \right])$ khi đó P$=\frac{t^{2}}{1+4t+t^{2}}$ ta cần cm $P\geq \frac{1}{6}$ $\Leftrightarrow 5t^{2}-4t-1\geq0 \Leftrightarrow (t-1)(5t+1)\geq0$ ( lđ vs $\forall t\in \left[ 1{;}4 \right]$)
Dấu "=" $\Leftrightarrow x=y=1;z=2$