Can Inequalities Be Proven? A Solution to a Complex Equation

[anemone]

Prove that $\sqrt{x^4+7x^3+x^2+7x}+3\sqrt{3x}\ge10x-x^2$.

 

[anemone]

Hint:

Spoiler

Try to think from the perspective of proving the inequality by dividing the domain into two or more intervals.

 

[anemone]

Solution of other:

Spoiler

Since $\sqrt{x^4+7x^3+x^2+7x}+3\sqrt{3x}\ge10x-x^2$ is true for $x\ge 10$, the proof can continue for the interval $x<10$.

Squaring both sides gives:

$2\cdot 3^{\dfrac{3}{2}}\sqrt{x}\sqrt{x^4+7x^3+x^2+7x}+x^4+7x^3+x^2+34x \ge x^4-20x^3+100x^2$$\dfrac{2\cdot 3^{\dfrac{3}{2}}\sqrt{x}\sqrt{x^4+7x^3+x^2+7x}}{\sqrt{x}}\ge -27x^2+99x-34$

Now we have 2 cases, one of which the RHS is positive, and one of which where it is negative.

Case I:

$0<x<\dfrac{-\sqrt{681}+33}{18}$ or $\dfrac{\sqrt{681}+33}{18}<x<10$ These give RHS a negative, so the relation holds.

Case II:
$\dfrac{-\sqrt{681}+33}{18}<x<\dfrac{\sqrt{681}+33}{18}$

RHS is positive in this interval, so squaring both sides again we have

$\left(\dfrac{2\cdot 3^{\dfrac{3}{2}}\sqrt{x}\sqrt{x^4+7x^3+x^2+7x}}{\sqrt{x}}\right) ^2 \ge (-27x^2+99x-34)^2$

$-729x^4+5454x^3-10881x^2+6840x-400\ge 0$

$-(3x-4)^2(81x^2-390x+25) \ge 0$ and this inequality holds true in the interval $\dfrac{-\sqrt{681}+33}{18}<x<\dfrac{\sqrt{681}+33}{18}$ and therefore, we can conclude that $\sqrt{x^4+7x^3+x^2+7x}+3\sqrt{3x}\ge10x-x^2$