\begin{eqnarray*}\sum\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}&=&\sum\frac{a^{2}+(a+b+c)^{2}}{2a^{2}+(b+c)^{2}}\\ &=&\sum\frac{2a^{2}+(b+c)^{2}+2a(b+c)}{2a^{2}+(b+c)^{2}}\\ &=& 3+\sum\frac{2a(b+c)}{2a^{2}+(b+c)^{2}}\\ &\leq& 3+\sum\frac{2a(b+c)}{2(ab+bc+ca)}=5\end{eqnarray*}

function f(x)=(x^2+9)/(2*x^2+(3-x)^2) is concave so f(a)+f(b)+f(c)<=3f((a+b+c)/3)=3f(1)=3*10/6=5

Good solution. I think it's better that you learn $\LaTeX$. It's easy.

bel.jad5 wrote: function f(x)=(x^2+9)/(2*x^2+(3-x)^2) is concave so f(a)+f(b)+f(c)<=3f((a+b+c)/3)=3f(1)=3*10/6=5 Also my solution. This is the first problem of the second day. And the Contest level is not very high, so there must exist simple and straightfoward solutions like that.

Since for $0<x<3,\ (x+4)\{2x^{2}+(3-x)^{2}\}-3(x^{2}+9)=(x-1)^{2}(x+3)\geq 0,$ we have $\sum\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}=\sum\frac{a^{2}+9}{2a^{2}+(3-a)^{2}}\leq \sum\frac{a+4}{3}=5.\ \ Q.E.D.$

bel.jad5 wrote: function f(x)=(x^2+9)/(2*x^2+(3-x)^2) is concave so f(a)+f(b)+f(c)<=3f((a+b+c)/3)=3f(1)=3*10/6=5 bel.jad5, how did you prove that $f(x)$ is concave?

Quote: bel.jad5, how did you prove that f is concave? just by mathematic intuition

When we use Jensen's inequality, we should show that $f(x)$ is concave up or down.

\[{a^{2}+9 \over 2a^{2}+(b+c)^{2}}={a^{2}+9 \over a^{2}-2a+3}\leq{1 \over 3}\left(1+{2a+6 \over (a-1)^{2}+2}\right) \leq{1 \over 3}(a+4).\]

bel.jad5 wrote: function $f(x)=\frac{x^{2}+9}{2x^{2}+(3-x)^{2}}$ is concave so $f(a)+f(b)+f(c)\leq 3f\left(\frac{a+b+c}{3}\right)= 3f(1)=3\cdot \frac{10}{6}= 5$ Are you sure that it is concave? $f''(x)= \frac{4\left(x^{3}+9x^{2}-27x+9\right)}{3\left(x^{2}-2x+3\right)^{3}}$. This expression is positive for $x\to 0$, because $f''$ is continuous in $0$ and $f''(0)=\frac49$, but negative for $x=1$: $f''(1)=\frac{-4}{3}$. Thus the function $f$ is neither convex nor concave on interval $(0,3)$.

Let $a,b,c$ are positive numbers, show that:\[\frac{a^2+9bc}{2a^2+(b+c)^2}+\frac{b^2+9ca}{2b^2+(c+a)^2}+\frac{c^2+9ab}{2c^2+(a+b)^2}\leq 5\]Let $a,b,c$ are positive numbers such that $a+b+c \geq 3$, show that:\[\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2b^{2}+(a+c)^{2}}+\frac{c^{2}+9}{2c^{2}+(a+b)^{2}}\leq 5\]\[\frac{a^{2}+9}{a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{b^{2}+(a+c)^{2}}+\frac{c^{2}+9}{c^{2}+(a+b)^{2}}\leq 6\]Let $a,b,c$ are positive numbers such that $abc \geq 1$, show that:\[\frac{a+9}{2a^{2}+(b+c)^{2}}+\frac{b+9}{2b^{2}+(a+c)^{2}}+\frac{c+9}{2c^{2}+(a+b)^{2}}\leq 5\]\[\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2b^{2}+(a+c)^{2}}+\frac{c^{2}+9}{2c^{2}+(a+b)^{2}}\leq 5\]\[\frac{a^{2}+4}{a^{2}+(b+c)^{2}}+\frac{b^{2}+4}{b^{2}+(a+c)^{2}}+\frac{c^{2}+4}{c^{2}+(a+b)^{2}}\leq 3\]Let $a,b,c\geq 0$. Prove that \[\frac{a^2}{2a^2+(b+c)^2}+\frac{b^2}{2b^2+(c+a)^2}+\frac{c^2}{2c^2+(a+b)^2}\leq \frac{2}{3}.\]

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