Inequality problem about sides of a triangle and the semiperimeter

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.

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inspite of voting it down...can you solve it...den vote it down @ Test123 –  soumajit das May 2 '14 at 3:38
This is not a homework help site. –  Awesome May 2 '14 at 3:39
@Test123 isn't necessarily the one who downvoted, and as far as I can tell, he's only the editor. –  2012ssohn May 2 '14 at 3:39
@Awesome - this may not be a site that was designed for homework help, but asking homework questions are fine. –  2012ssohn May 2 '14 at 3:40
On that note, @soumajitdas, I would recommend providing some more detail about what you have tried so far, and where exactly you're stuck. –  2012ssohn May 2 '14 at 3:41

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

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

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 $a=2b=2c$.

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but its 36/35 @ calvin lin. –  soumajit das May 2 '14 at 5:10
@soumajitdas he has proved your case first, and $36/35 < 6/5$, so the last one is stronger. –  Macavity May 2 '14 at 5:31
@soumajit note that I only proved 36/35 and not 6/5. The latter is easily deduced using my method. –  Calvin Lin May 2 '14 at 6:27