Prove that $x^2 + y^2+ z^2\le xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

Prove that $x^2 + y^2+ z^2\le  xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

For all $x,\,y,\, z\in [0,1]$, we know $x^2 + y^2+ z^2\le x+y+z$.

If one root of $4x^2+2x-1= 0$ be $\alpha$, please show that other root is $4\alpha^3-3\alpha$.

If one root of $4x^2+2x-1= 0$ be $\alpha$, please show that other root is $4\alpha^3-3\alpha$.

My solution:

The relation of $2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$.

Prove that if in a triangle $ABC$ we have the following equality that holds

$2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$

then the triangle will be an equilateral triangle.

In any triangle $ABC$, we have the following equality that holds:

Compare which of the following is bigger: $\displaystyle 1016^{11}\cdot 3016^{31}$ versus $\displaystyle 2016^{42}$

Compare which of the following is bigger:

$\displaystyle 1016^{11}\cdot 3016^{31}$ versus $\displaystyle 2016^{42}$

My solution:

Prove that : $\displaystyle \frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}$ for all positive reals $a$ and $b$.

Prove that :
$\displaystyle \frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}$ for all positive reals $a$ and $b$.

My solution:

Step 1:

Solve for real solution(s) for $\displaystyle x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$.

Solve for real solution(s) for $\displaystyle x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$.

My solution: