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.

Answer 2

$\Rightarrow$ We will consider triangle $OAB$ wher $A,\ B$ are vectors on $\mathbb{R}^2$ Define $$a=|A|,\ b=|B|,\ c=|A-B|$$ Then $$ |A|^2+|B|^2-2A\cdot B=|A-B|^2 $$

Then $pA+qB,\ p+q=1$ is in a line passing through $A,\ B$. Then \begin{align} |pA+qB|^2&=p^2a^2+q^2b^2+2pqA\cdot B \\&= p^2a^2+q^2b^2+pq (a^2+b^2-c^2) \\& =pa^2+ qb^2 -pqc^2 \ \ast\end{align}

Since $a,\ b>0$, so $|pA+qB|>0$ Hence "only if" is proved

$\Leftarrow$ In $\ast$, $$pa^2+ qb^2 -pqc^2=(pa)^2+ (qb)^2-2(pa)(pb) C $$ where $$ C= - \frac{a^2+b^2-c^2}{2ab }$$

If $C>1$ then $c>a+b $ So for $pa=qb$ we have $pa^2+qb^2 > pqc^2>pq(a+b)^2\Rightarrow (pa-qb)^2>0 $ It is a contradiction

If $C<-1$ then $|a-b|>c$ If $a>b$ then $a>b+c$ so that for $ p=-t,\ t>0,\ q=1+t$ we have $ (b-tc)^2>0$ Let $t=\frac{b}{c}$ Hence it is a contradiction

So $|C|\leq 1$ So we have $\theta$ s.t. $\cos\ \theta=-C$ If we have triangle of sides $a,\ b,\ c'$ where $\theta$ is an angle between sides of length $a,\ b$, then $\cos\ \theta=(a^2+b^2-(c')^2)/2ab$ That is $c=c'$ This complete the proof