3 Applications of the Cauchy Schwarz

From the Cauchy- Schwarz inequality, we have $(x^2+y^2+z^2)^2=3(x^2+y^2+z^2) \geq (x+y+z)^2.$ This gives, $\boxed{x^2+y^2+z^2 \geq x+y+z}.$ Again using the Cauchy- Schwarz inequality, we have $(x^2+y+z)(1+y+z) \ge (x+y+z)^2,$ and their symmetric equivalents. Hence, it suffices to prove that $\frac{x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y}}{x+y+z} \leq \sqrt{3}.$ Now using Cauchy- Schwarz inequality for the third time, we have $(x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y})^2=(\sqrt{x}\sqrt{x+xy+xz}+\sqrt{y}\sqrt{y+yx+yz}+\sqrt{z}\sqrt{z+zx+zy})^2 \leq (x+y+z)(x+y+z+2xy+2yz+2zx) \leq (x+y+z)(x^2+y^2+z^2+2xy+2yz+2zx)=(x+y+z)^3.$ This gives, $\frac{x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y}}{x+y+z} \leq \sqrt{x+y+z} \leq \sqrt{x^2+y^2+z^2}=\sqrt{3}.$ Hence, the square of the maximum value of the wanted expression is $3.$ Note that the equality occurs at $x=y=z=1$ .