Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}1+x\ge2\sqrt{x}\\x+y\ge2\sqrt{xy}\\1+y\ge2\sqrt{y}\end{matrix}\right.\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right)\)
\(\Leftrightarrow VT=1+x+y\ge\sqrt{x}+\sqrt{y}+\sqrt{xy}=VP\)
Xảy ra khi \(\left\{{}\begin{matrix}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\1+y=2\sqrt{y}\end{matrix}\right.\)\(\Rightarrow x=y=1\)
Khi đó \(P=x^2+y^2=1^2+1^2=2\)
Và \(Q=x^{2009}+y^{2009}=1^{2009}+1^{2009}=2\)
Với \(x,y>0\) ta có
\(1+x+y=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
\(\Leftrightarrow2+2x+2y-2\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)+\left(x-2\sqrt{xy}+y\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2=0\)
\(\forall x,y>0\) ta luôn có \(\left\{{}\begin{matrix}\left(\sqrt{x}-1\right)^2\ge0\\\left(\sqrt{y}-1\right)^2\ge0\\\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=1\)
Vậy x=y=1
Nên P=Q=2