Prove the Condition for Construction of a Triangle Let $a,b,c>0$ be real numbers. Show that you can construct the triangle with length sides $a,b,c$ if and only if $pa^2+qb^2>pqc^2$, for any $p,q$ such that $p+q=1$.
 A: As Exodd said, the statement is false. Let $f_{a,b,c}(p)=pa^2+2b^2-p(1-p)c^2=c^2p^2+(a^2-c^2)p+2b^2$. Then your statement says: $a,b,c>0$ are the sides of a triangle if, and only if, $f_{a,b,c}>0$.
Take $a=c=1$, $b=3$. Then $f_{a,b,c}(p)=p^2+6$, which is always positive, but $(1,1,3)$ cannot be the three sides of a triangle.

If we replace $2b^2$ by $qb^2$, then we get a much more interesting problem that we can prove. Once again, define
$$\begin{aligned}
f_{a,b,c}(p)&=pa^2+(1-p)b^2-p(1-p)c^2 \\
&=c^2p^2+(a^2-b^2-c^2)p+b^2,
\end{aligned}$$
and the problem becomes: $a,b,c>0$ are the sides of a triangle if, and only if, $f_{a,b,c}>0$. Moreover, we can suppose $a<b$, since their roles are symmetrical.
Since the leading coefficient $c^2$ of the quadratic function $f_{a,b,c}$ is positive, we have
$$\begin{aligned}
f_{a,b,c}>0&\Leftrightarrow \Delta=(a^2-b^2-c^2)^2-4b^2c^2<0\\
&\Leftrightarrow (a^2-b^2-c^2)^2<4b^2c^2 \\
&\Leftrightarrow 0<b^2+c^2-a^2<2bc \\
&\Leftrightarrow b^2-2bc+c^2=(b-c)^2<a^2 \\
&\Leftrightarrow |b-c|<a.
\end{aligned}$$
And by hypothesis, $a<b<b+c$, so we have
$$f_{a,b,c}>0\Leftrightarrow |b-c|<a<b+c,$$
and the right hand side is the triangle inequality.
A: The statement is false, since you can take $b$ as big as you want. 
For example, if $a=c=1$, and $b=3$, the inequality surely holds, but there's no triangle with such sides length.
