# Finding the smallest $k$ such that $a^3+b^3+c^3+kabc\leq\frac16(k+3)(a^2(b+c)+b^2(a+c)+c^2(a+b))$, where $a,b,c$ are sides of a triangle

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

## 1 Answer

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.

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