Prove $\sum\frac{a}{(2k^2-3k+2)a+k^2b+c}\le \frac{1}{k^2-k+1}$

Question

Question

If $a,b,c\ge 0:a+b+c>0$ and non-negative constant $k$ then prove$$\frac{a}{(2k^2-3k+2)a+k^2b+c}+\frac{b}{(2k^2-3k+2)b+k^2c+a}+\frac{c}{(2k^2-3k+2)c+k^2a+b}\le \frac{1}{k^2-k+1}$$ Equality cases: $(1,1,1);(k,1,0)$ and cyclic permutations.

It is true after verifying by Mathematica. I am waiting nice proofs.

For each $k$ we obtain nice consequenced problems.

• $k=0$ is a special case $$\frac{a}{2a+c}+\frac{b}{2b+a}+\frac{c}{2c+b} \le 1$$which is easy by C-S. Equality occurs at $a=b=c>0.$

• $k= 1:$ we get an equality.

• $k=2$ is also an old problem $$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le \frac{1}{3}$$ Equality occurs at $(t,t,t);(2t,t,0),t>0$ and cyclic permutations.

• $k=3$ is asked here $$\frac{a}{11a+9b+c}+\frac{b}{11b+9c+a}+\frac{c}{11c+9a+b}\le \frac{1}{7}$$Equality occurs at $(t,t,t);(3t,t,0),t>0$ and cyclic permutations.

All idea is welcome. Thanks.

Answer 1

Short proof with a computer :

Setting $k=C\leq 1$ as particular value then :

$$p(a,b,c)=\frac{a}{(2k^2-3k+2)a+k^2b+c}+\frac{b}{(2k^2-3k+2)b+k^2c+a}+\frac{c}{(2k^2-3k+2)c+k^2a+b}-\frac{1}{k^2-k+1}$$

Then :

$$p(ka+b+c,b+c,b)\leq 0$$

Because all the coefficient are negatives .

Example :

$k=0.6$ :

$$-100(3249ba^{2}+5415bac+9025bc^{2}+1215a^{3}+9270a^{2}c+10740ac^{2}+3400c^{3})/(\operatorname{positivedenominator)}\leq 0$$

We can do the same with $p(ka+b+c,b,b+c),k\geq 1$ and $p(ka+b+c,a+b+c,c),k\in[0,\infty]$

We are done

Answer 2

After full expanding we need to prove that: $$k^4\sum_{cyc}(a^2c-abc)-2k^3\sum_{cyc}(a^2b-abc)+k^2\sum_{cyc}(a^3-abc)-2k\sum_{cyc}(a^2c-abc)+\sum_{cyc}(a^2b-abc)\geq0.$$ Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Thus, we need to prove that: $$k^4((u^2-uv+v^2)a+uv^2)-2k^3((u^2-uv+v^2)a+u^2v)+k^2(3(u^2-uv+v^2)a+u^3+v^3)-2k((u^2-uv+v^2)a+uv^2)+(u^2-uv+v^2)a+u^2v\geq0$$ or $$(u^2-uv+v^2)a(k^4-2k^3+3k^2-2k+1)+k^2u^3-(2k^3-1)uv^2+(k^4-2k)uv^2+k^2v^3\geq0$$ and since $$k^4-2k^3+3k^2-2k+1>0,$$ it's enough to prove that:$$k^2u^3-(2k^3-1)u^2v+(k^4-2k)uv^2+k^2v^3\geq0$$ or $$(k^2u+v)(u-kv)^2\geq0$$ and we are done.