Inequality problem about sides of a triangle and the semiperimeter

Question

Let $a,b,c$ the sides of a triangle and $s$ be the semi perimeter. Then show that $$ a^2+b^2+c^2 > \frac{36}{35}(a^2+\frac{abc}{s}) $$ I tried it doing in many ways using some changes but cannot help my cause.

Answer 1

The inequality seems really loose to me, and this approach shows that it is indeed loose. ( which makes me somewhat doubt the inequality is what we want to show)

We want to show that

$$ 35 b^2 +35 c^2 > a^2 + 36 abc/s .$$

We have $b+c > a$ and so $ 2s > 2a $ and so $abc/s < bc $. We will show that

$$ 35 b^2 +35 c^2 > a^2 + 36 bc .$$

This is true because $a < b+c $ so $a^2 < 2 b^2 + 2c^2$, which gives us

$$ 18(b-c)^2 +17 b^2 +17 c^2 > a^2$$

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In fact, with the above, we can show that

$$ a^2+b^2+c^2 > \frac{6}{5}(a^2+\frac{abc}{s}) $$

the 'equality condition' occurs when $a=2b=2c$.