<s>[INST] Show that $x^2 + y^2 + 1 \le \sqrt{(x^3 + y + 1)(y^3 + x + 1)}$ For $x, y \ge 0$ prove that:
$$x^2 + y^2 + 1 \le \sqrt{(x^3 + y + 1)(y^3 + x + 1)}$$
What I think would apply is the AM-GM Inequality, so first,
$$(x^2 + y^2 + 1)^2 \le (x^3  + y + 1)(y^3 + x + 1)$$
$$\implies (x^3  + y + 1)(y^3 + x + 1) - (x^2 + y^2  + 1)^2 \ge 0$$
$$\implies x^3y^3 + x^3 - 2x^2y^2 - 2x^2 + xy + x + y^3 - 2y^2 + y \ge 0$$
From AM-GM,
$$\frac{ x^3y^3 + x^3 - 2x^2y^2 - 2x^2 + xy + x + y^3 - 2y^2 + y}{9} \ge \sqrt[9]{x^3y^3(x^3)(-2x^2y^2) (-2x^2)(xy) (x) (y^3) (- 2y^2) (y)}$$ 
The negative is the issue we have: $(-2x^2)(-2x^2y^2)(-2y^2)$ which is a problem.
How can this be done? Please, HINTS only.
 [/INST] Solution 1:
Only use Cauchy-Schwarz inequality :
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$
let $$a_{1}=x^{\frac{3}{2}},a_{2}=y^{\frac{1}{2}},a_{3}=1$$
$$b_{1}=x^{\frac{1}{2}},b_{2}=y^{\frac{3}{2}},b_{3}=1$$
so we have
$$(x^3+y+1)(x+y^3+1)\ge (x^2+y^2+1)^2$$
so
$$\sqrt{(x^3+y+1)(y^3+x+1)}\ge x^2+y^2+1$$
Solution 2:
you can also 
$$\Longleftrightarrow (x^3y^3-2x^2y^2+xy)+(x^3-2x^2+x)+(y^3-2y^2+y)\ge 0$$
since use AM-GM inequality
$$ x^3y^3+xy\ge 2\sqrt{x^3y^3\cdot xy}=2x^2y^2$$
$$x^3+x\ge 2\sqrt{x^3\cdot x}=2x^2$$
$$y^3+y\ge 2\sqrt{y^3\cdot y}=2y^2$$
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