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Let $a,b,c\ge 0$, and assume $a+b+c=3$.
I'd like to show that $$\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}$$

My try: Use Hölder inequality $$\left(\sum_{cyc}\dfrac{a}{\sqrt{b^2+b+1}}\right)\left(a\sqrt{b^2+b+1}\right)^2\ge (a+b+c)^3\,,$$ then we only need to prove this $$\sum_{cyc}a(b^2+b+1)=\sum_{cyc}(ab^2+ab+a)\le 9$$ but $$ab^2+bc^2+ca^2+ab+bc+ac\le 6$$ is not true.

So I can't prove it.

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  • $\begingroup$ No,This can't use dearrangement inequality.and $\dfrac{a}{\sqrt{a^2+a+1}}\ge\dfrac{1}{3}$ is not true $\endgroup$
    – math110
    Feb 16, 2014 at 17:21
  • $\begingroup$ Hello,$\dfrac{1}{\sqrt{3}}$ is also can't true,such $a=0.00000001$ $\endgroup$
    – math110
    Feb 16, 2014 at 17:29

1 Answer 1

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I'm going to use two facts. The first one can be obtained as following: $$\begin{align} ab + bc + ca &\leqslant a^2 + b^2 + c^2, \\ 3(ab + bc + ca) &\leqslant (a + b + c)^2 = 9, \\ ab + bc + ca &\leqslant 3. \end{align}$$

And the second one: $$\frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}}$$ (here's Wolfram Alpha visualization, a bit later I'm going to prove it more rigorously).

So, putting it all together: $$ \begin{align} \frac{a}{\sqrt{b^2 + b + 1}} + \frac{b}{\sqrt{c^2 + c + 1}} + \frac{c}{\sqrt{a^2 + a + 1}} &\geqslant \frac{\sqrt{3}}{2}(a + b + c) - \frac{ab + bc + ca}{2\sqrt{3}} \\ &\geqslant \frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \sqrt{3} \end{align}$$


Lemma. $\displaystyle \frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}},\ \forall x \geqslant 0$.

Note that $\displaystyle \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}(3 - x) < 0$ for $x > 3$. So, for these $x$ we get obvious inequality that square root of something is greater than $0$.

For $x \in [0, 3]$: $$ \begin{align} \frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}} &\Longleftrightarrow 12 \geqslant (x^2 + x + 1)(3 - x)^2 \\ &\Longleftrightarrow x^4 - 5x^3 + 4x^2 +3x - 3 \leqslant 0 \\ &\Longleftrightarrow (x - 1)(x^3 - 4x^2 + 3) \leqslant 0 \\ &\Longleftrightarrow (x - 1)^2(x^2 - 3x - 3) \leqslant 0 \end{align}$$ which is true because roots of $x^2 - 3x - 3$ are: $x_1 = \frac{1}{2}(3 - \sqrt{21}) < 0$, $x_2 = \frac{1}{2}(3 + \sqrt{21}) > 3$. So for $x \in [0, 3]$ holds $x^2 - 3x - 3 < 0.$

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    $\begingroup$ nice!for last one , it is trivial :$x^2-3x-3=x(x-3)-3<0$ $\endgroup$
    – chenbai
    Feb 17, 2014 at 23:59
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    $\begingroup$ Geometric content of the lemma: The LHS function is (strictly) convex, and RHS is the tangent to it in $(1, \sqrt 3/3)$. $\endgroup$
    – Hanno
    Feb 10, 2021 at 17:30
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    $\begingroup$ Continued comment / specification / nitpicking: At $\,(\sqrt 6 -2)/4\approx 0.112\,$ the LHS function has an point of inflexion. It is strictly convex only from thereon towards $\infty$. Nevertheless, the tangent estimate is valid for all $x\geqslant 0$, as needed. $\endgroup$
    – Hanno
    Feb 11, 2021 at 13:39

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