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

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: $a + b\geq c,; b + c\geq a,\; a + c\geq b$.

A: Let two of the sides be $a$ and $b$, and let the angle between them be $\theta$. 
Then the area is $\frac{1}{2}ab\sin\theta$, so $16$ times that is $8ab\sin\theta$.
By the Cosine Law, the square $c^2$ of the third side is $a^2+b^2-2ab\cos\theta$. It follows that 
$$2a^2+2b^2+3c^2=5a^2+5b^2-6ab\cos\theta.$$
We are interested in proving the inequality 
$$5a^2+5b^2-6ab\cos\theta\overset{?}{\ge} 8ab\sin\theta,$$
or equivalently 
$$5a^2-2ab(3\cos\theta+4\sin\theta)+5b^2 \overset{?}{\ge} 0.$$
We would like to show that the discriminant is non-positive. So we would like to show that
$$(3\cos\theta+4\sin\theta)^2 \overset{?}{\le} 25.$$
This comes down to computing the maximum of $|3\cos\theta+4\sin\theta|$. But that maximum is $5$. There are many ways to show this. For example, let $\phi$ be the angle whose sine is $\frac{3}{5}$ and whose cosine is $\frac{4}{5}$. Then
$3\cos\theta+4\sin\theta=5\sin(\theta+\phi)$. 
Remark: We have equality when $a=b=\sqrt{5}$ and $c=2$. 
A: Here is a short algebraic proof.  Since both sides are homogeneous of degree $2$, we can rescale to the case where $c=1$.  Then  by translating $A$ to the origin and by rotating so that $B$ lies at $(1,0))$ , we can reduce to the case where $A=(0,0)$, $B=(1,0)$ and $C$ is $(p,q)$ for some $p$ and $q$.  The inequality then becomes
$$2(p^2+q^2) + 2((p-1))^2 + q^2) + 3 \geq 16 q.$$
This is,  by simple algebra, equivalent to the valid inequality 
$ (p-\frac 12 )^2+(q-1)^2 \geq 0$.
