Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$

Question

Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$ for any $p,q$ with $p+q = 1$.

Should I prove this using the triangle inequality? Also should I split the proof up into two parts for the if and only if?

Answer 1

Below is a complete solution. Don't look at it if you want to solve this problem by yourself. Also, yes, the Triangle Inequality is definitely required in this problem.

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Assume that $a,b,c>0$ satisfy the condition that $pa^2+qb^2>pqc^2$ for any $p,q \in\mathbb{R}$ such that $p+q=1$. Choosing $p$ and $q$ to be $\frac{b}{a+b}$ and $\frac{a}{a+b}$, respectively, we obtain $c<a+b$. Choosing $p$ and $q$ to be $\frac{b}{b-a}$ and $\frac{a}{a-b}$, respectively, we obtain $c>|a-b|$, which implies $a<b+c$ and $b<c+a$. Ergo, $a$, $b$, and $c$ form a triangle.

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Conversely, suppose that $a,b,c>0$ form a triangle. Let $p,q\in\mathbb{R}$ be such that $p+q=1$. If $p=0$ or $q=0$, then the inequality $pa^2+qb^2>pqc^2$ holds trivially. Now, suppose that $p,q>0$. Then, by the Cauchy-Schwarz Inequality, $$\frac{a^2}{q}+\frac{b^2}{p}=\left(\frac{a^2}{q}+\frac{b^2}{p}\right)(q+p)\geq (a+b)^2>c^2\,,$$ whence $pa^2+qb^2>pqc^2$ holds. If $p>0$ and $q<0$, then we note by symmetry that $\frac{1}{1-q}a^2+\frac{-q}{1-q}c^2>\frac{(-q)}{(1-q)^2}b^2$, which gives $$pa^2-pqc^2>\frac{p(-q)}{1-q}b^2=-qb^2\,,$$ as $p=1-q$. The previous inequality is precisely $pa^2+qb^2>pqc^2$. If $p<0$ and $q>0$, we can prove the inequality $pa^2+qb^2>pqc^2$ in a similar manner.