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Is it true that if $a+b+c=1$, where $a$, $b$ and $c$ are nonnegative real numbers, then $$(3-2\sqrt{2})\sum\limits_{cyc}\sqrt{ab}+2\sqrt{2}-1\geq\sum\limits_{cyc}\sqrt{(1-a)(1-b)}?$$ Edit: I was told to improve this question in a way that I assume means providing context. The similar inequality $\sum\limits_{cyc}\sqrt{(1-a)(1-b)}\geq\sum\limits_{cyc}\sqrt{ab}+1$ under the same conditions is an old and rather easy question. And since $\sum\limits_{cyc}\sqrt{ab}\leq1$, the inequality strengthens as the constant increases. It is easy to show that 1 and 1 are the best possible constants for a lower bound. Also 2 is an upper bound for the LHS of the original, so a natural question is how much can we lower this number? Playing around with it for a while strongly suggests that $(\frac{1}{2},\frac{1}{2},0)$ will be an equality case, giving the bound of $3-2\sqrt{2}$ for the coefficient of the cyclic sum term as shown. Unfortunately, due to my inexperience in more advanced or creative techniques in inequalities, this seems to be well out of my reach.

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  • $\begingroup$ where is $c$? what means $cyc$? $\endgroup$ – Masacroso Mar 8 '17 at 13:19
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We'll replace $a$ with $a^2$, $b$ with $b^2$ and $c$ with $c^2$.

Thus, $a^2+b^2+c^2=1$ and we need to prove that $$\sum_{cyc}\sqrt{(a^2+b^2)(a^2+c^2)}\leq\sum_{cyc}((2\sqrt2-1)a^2+(3-2\sqrt2)ab).$$ Now, by C-S $$\left(\sum_{cyc}\sqrt{(a^2+b^2)(a^2+c^2)}\right)^2\leq\sum_{cyc}\frac{(a^2+b^2)(a^2+c^2)}{(2\sqrt2-1)a+b+c}\sum_{cyc}((2\sqrt2-1)a+b+c).$$ Id est, it's enough to prove that $$\left(\sum_{cyc}((2\sqrt2-1)a^2+(3-2\sqrt2)ab)\right)^2\geq(2\sqrt2+1)(a+b+c)\sum_{cyc}\frac{(a^2+b^2)(a^2+c^2)}{(2\sqrt2-1)a+b+c}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

We see that the left side of the last inequality does not depend on $w^3$;

The denominator of the right side it's $$\prod_{cyc}((2\sqrt2-1)a+b+c),$$ which is a linear expression of $w^3$ and the nominator of the right side it's $$(2\sqrt2+1)(a+b+c)\sum_{cyc}(a^2+b^2)(a^2+c^2)((2\sqrt2-1)b+a+c)((2\sqrt2-1)c+a+b)=$$ $$=(2\sqrt2+1)(a+b+c)\sum_{cyc}(a^4+a^2b^2+a^2c^2+b^2c^2)((2\sqrt2-1)b+a+c)((2\sqrt2-1)c+a+b)=$$ $$=3(2\sqrt2+1)u\sum_{cyc}(a^4+9v^4-6uw^3)((2\sqrt2-2)b+3u)((2\sqrt2-2)c+3u).$$ The expression $$3(2\sqrt2+1)u\sum_{cyc}(9v^4-6uw^3)((2\sqrt2-2)b+3u)((2\sqrt2-2)c+3u)$$ is a linear expression of $w^3$.

Thus, we can get a non-zero coefficient before $w^6$ in the following expression only. $$3(2\sqrt2+1)u\sum_{cyc}a^4(2\sqrt2-2)b(2\sqrt2-2)c=3(2\sqrt2+1)(2\sqrt2-2)^2uw^3(a^3+b^3+c^3)=$$ $$=3(2\sqrt2+1)(2\sqrt2-2)^2uw^3(27u^3-27uv^2+3w^3)$$ and since $$3(2\sqrt2+1)(2\sqrt2-2)^2\cdot3>0,$$ we see that we need to prove that $f(w^3)\leq0,$ where $f$ is a convex function.

But the convex function gets a maximal value for an extreme value of $w^3$,

which happens in the following cases.

  1. $w^3=0$.

Since the equality $$\left(\sum_{cyc}((2\sqrt2-1)a^2+(3-2\sqrt2)ab)\right)^2\geq(2\sqrt2+1)(a+b+c)\sum_{cyc}\frac{(a^2+b^2)(a^2+c^2)}{(2\sqrt2-1)a+b+c}$$ is homogeneous and symmetric,

it's enough to assume $c=0$ and $b=1$, which gives

$$((2\sqrt2-1)(a^2+1)+(3-2\sqrt2)a)^2\geq$$ $$\geq(2\sqrt2+1)(a+1)\left(\frac{(a^2+1)a^2}{(2\sqrt2-1)a+1}+\frac{a^2+1}{a+2\sqrt2-1}+\frac{a^2}{a+1}\right)$$ or $$(a-1)^2((10\sqrt2-13)a^4+(74-49\sqrt2)a^3+(71\sqrt2-94)a^2+(74-49\sqrt2)a+10\sqrt2-13)\geq0,$$ which is obvious;

  1. Two variables are equal.

Let $b=c=1$.

We need to prove that: $$\left((2\sqrt2-1)(a^2+2)+(3-2\sqrt2)(2a+1)\right)^2\geq(2\sqrt2+1)(a+2)\left(\frac{(a^2+1)^2}{(2\sqrt2-1)a+2}+\frac{4(a^2+1}{a+2\sqrt2}\right)$$ or $$(a-1)^2a((10\sqrt2-13)a^3+36(3-2\sqrt2)a^2+(170\sqrt2-231)a+64-40\sqrt2)\geq0,$$ which is obvious again and we are done!

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