Prove or disprove inequality: $2a^2 + 2b^2 + 3c^2 \ge 16P$ Let $a,b,c$ be the lengths of the sides of a triangle with area $P$ 
prove or disprove inequality:  $2a^2 + 2b^2 + 3c^2 \ge 16P$


$$2a^2 + 2b^2 + 3c^2 \overset{?}{\ge} 16P\tag{1}$$

Hint: It will be helpful to know Heron's formula that provides a formula for the area of a triangle with sides length $a, b, c$:
$$P = \sqrt{s(s-a)(s-b)(s-c)}$$
where $ s=\dfrac{a+b+c}{2}$. This gives us
$$\text{Area}\;=\;\;P = \frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}$$
where $a, b, c$ are the side lengths of a triangle
Substituting $P$ into your original inequality, you can work with proving/disproving the following inequality: 
$$2a^2 + 2b^2 + 3c^2 \geq 4\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}\tag{2}$$


*

*Start with squaring each side of the inequality $(2)$. (This is fine to do, without needing to consider a change in the direction of the inequality, because both sides are non negative.)
$$(2a^2 + 2c^2 + 3c^2)^2 \geq 16(a^2 + b^2 + c^2)^2 - 32(a^4 + b^4 + c^4)\tag{3}$$

*If you need it, you also have the triangle inequality to work with to put additional restrictions on the inequality: $a + b\geq c,; b + c\geq a,\; a + c\geq b$.

