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If $ a^2 + b^2 > 5c^2 $ in a triangle ABC then show c is the smallest side.I tried to solve this by cosine rule but i was not able to find the answer

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First, recall that in any triangle, the triangle inequality states that $a+b>c$, $b+c>a$, and $c+a>b$.

Combining this with your condition $a^2+b^2>5c^2$, we have

$$5c^2<a^2+b^2<(b+c)^2+b^2=2b^2+2bc+c^2,$$

hence $$2c^2<b^2+bc.$$

Now, suppose it were the case that $b\leq c$. Then $2c^2<b^2+bc\leq 2c^2$, a contradiction.

A similar argument holds when $a\leq c$, hence $c$ is indeed the smallest side.

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Suppose WLOG $c> a$. Then $c^2+b^2> a^2+b^2> 5c^2 \implies b^2> 4c^2 \implies b > 2c > c+a$ which would violate the triangle inequality.

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