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:  $2a^2 + 2b^2 + 3c^2 \ge 16P$

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 $c=2$. 
