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Let $a,b,c \in \mathbb{R_+}$. Prove that
$$\frac{a^2}{b^2+c^2} + \frac{b^2}{c^2+a^2} + \frac{c^2}{a^2 + b^2} \geq \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}. $$

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  • $\begingroup$ why so more (-1) votes? Inequality seems to be interesting :) $\endgroup$ – Iuli Sep 18 '12 at 20:48
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Note that

\begin{equation} \begin{split} &\dfrac{a^2}{b^2 + c^2} - \dfrac{a}{b+c} = \dfrac{ab(a-b) + ac(a-c)}{(b+c)(b^2 + c^2)}, \\ &\dfrac{b^2}{c^2 + a^2} - \dfrac{b}{c+a} = \dfrac{bc(b-c) + ab(b-a)}{(c+a)(c^2 + a^2)}, \\ &\dfrac{c^2}{a^2 + b^2} - \dfrac{c}{a+b} = \dfrac{ac(c-a) + bc(c-b)}{(b+a)(b^2 + a^2)}. \end{split} \end{equation}

Thus we can rewrite the inequality we are to prove as:

\[ \dfrac{ab(a-b) + ac(a-c)}{(b+c)(b^2 + c^2)} + \dfrac{bc(b-c) + ab(b-a)}{(c+a)(c^2 + a^2)} + \dfrac{ac(c-a) + bc(c-b)}{(b+a)(b^2 + a^2)} \geq 0.\]

The left-hand side equals

\begin{equation} \begin{split} &ab(a-b) \cdot \left( \dfrac{1}{(b+c)(b^2 + c^2)} - \dfrac{1}{(a+c)(c^2 + a^2)} \right) + \\ &ac(a-c) \cdot \left( \dfrac{1}{(b+c)(b^2 + c^2)} - \dfrac{1}{(b+a)(b^2 + a^2)} \right) + \\ &bc(b-c) \cdot \left( \dfrac{1}{(c+a)(c^2 + a^2)} - \dfrac{1}{(b+a)(b^2 + a^2)} \right). \end{split} \end{equation}

Now let's take the expression in the first bracket:

\begin{equation} \begin{split} &\dfrac{1}{(b+c)(b^2 + c^2)} - \dfrac{1}{(a+c)(c^2 + a^2)} = \dfrac{(a+c)(c^2 + a^2) - (b+c)(b^2 + c^2)}{(b+c)(b^2 + c^2)(a+c)(c^2+a^2)}. \end{split} \end{equation}

The numerator can be expanded as follows:

\begin{equation} \begin{split} &(a+c)(c^2 + a^2) - (b+c)(b^2 + c^2) = (a-b)(a^2 + b^2 + c^2 + ab) + c(a^2 - b^2) = \\ &(a-b)(a^2 + b^2 + c^2 + ab + c(a+b)). \end{split} \end{equation}

So the first term equals \[ (a^2 + b^2 + c^2 + ab + bc + ca) \cdot \dfrac{ab(a-b)^2}{(b+c)(b^2 + c^2)(a+c)(c^2+a^2)}. \]

The other terms can be transformed by analogy. Finally, we obtain

\[ (a^2 + b^2 + c^2 + ab + bc + ca) \cdot \sum_{cyc} \dfrac{ab(a-b)^2}{(b+c)(b^2 + c^2)(a+c)(c^2+a^2)}, \]

which obviously is nonnegative for all $a, b, c \in \mathbb{R_+}$.

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Notice that when you square terms like:

$\left(\frac{a}{b+c}\right)^2=\frac{a^2}{b^2+2bc+c^2}\leq \frac{a^2}{b^2+c^2}$

for $c>0$. So your ineqality is true for $c>0$. The inequality you wrote is false for $c<0$ (take $c$ close to $-a$ for example).

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