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It is known that $a, b, c$ are the sides of the triangle. Determine the smallest value of $k$, so that $$a^3+b^3+c^3+kabc\leq\frac {k+3}{6} (a^2(b+c)+b^2(a+c)+c^2(a+b))$$

My working: $$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (a^2(b+c)+b^2(a+c)+c^2(a+b))$$ $$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)$$ $$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (a^2b+b^2a+a^2c+c^2a+b^2c+c^2b)$$ $$a^3+b^3+c^3+kabc≤ \frac {k+3}{6} (ab(a+b)+ac(a+c)+bc(b+c))$$

Can someone help me, I only process the data on the right side? Thank you

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1 Answer 1

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Since $a,b$ and $c$ are sides of a triangle, we are allowed to set $a=x+y$, $b=y+z$ and $c=x+z$. Now, let's define: $$A=x^3+y^3+z^3 \\ B=x^2y+xy^2+z^2x+zx^2+y^2z+yz^2 \\C=xyz.$$

Note that by the Schur's inequality we already know that: $A+3C\ge B.$ Moreover it is not hard to see (by Muirhead's inequality) that $2A\ge B.$

Then, writing the inequality in terms of $A,B$ and $C$, we should have: $$2A+3B+k(B+2C) \le \frac{k+3}{6} (2A+5B+12C). $$

By simplifying, we should have:

$$(\frac {k+3}{6})B\le (\frac {k-3}{3})A+6C.$$

If $k\ge 9$, then:

$$(\frac {k-3}{3})A+6C=2A+6C+(\frac{k-9}{3})A\ge 2B+ (\frac{k-9}{6})B=(\frac{k+3}{6})B.$$

Therefore the inequality holds if $k\ge9$.


Now, assume $k\lt 9$; then $\frac {k+3}{6} \gt \frac {k-3}{3}.$ Let's put $x=1, y=1, z=\epsilon$.

We should have:

$$(\frac{k+3}{6})(2+2\epsilon^2+2\epsilon)\le (\frac{k-3}{3})(2+\epsilon^3)+6\epsilon,$$ which is impossible for $\epsilon$ small enough.

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  • $\begingroup$ You are right. Thank you! $\endgroup$ Dec 17, 2022 at 10:34

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