\[\large{\frac { ab+1 }{ { ( a+b ) }^{ 2 } } +\frac { bc+1 }{ ( b+c ) ^{ 2 } } +\frac { ca+1 }{ { ( c+a ) }^{ 2 } } \ge 3}\]

Let \(a,b\) and \(c\) be positive real numbers such that \(a^2+b^2+c^2+(a+b+c)^2 \le 4\). Prove the above inequality.

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TopNewestHere is my solution for the problem:

Since

\[ \begin{align*} (a+b)^2 + (b+c)^2 + (c+a)^2 &= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\ &= a^2 + b^2 + c^2 + (a + b + c)^2, \end{align*}\]

Let

\[ \begin{align*} \alpha &= b + c \\ \beta &= c + a \\ \gamma &= a + b \end{align*}\]

This implies

\[\begin{align*} a &= \frac{\beta + \gamma - \alpha}2 \\ b &= \frac{\alpha + \gamma - \beta}2 \\ c &= \frac{\alpha + \beta - \gamma}2 \end{align*}\]

With this change of variables, the constraint becomes

\[\alpha^2 + \beta^2 + \gamma^2 \le 4,\]

while the left side of the inequality we need to prove is now

\[\begin{align*} & \frac{\gamma^2 - (\alpha - \beta)^2 + 4}{4\gamma^2} + \frac{\alpha^2 - (\beta - \gamma)^2 + 4}{4\alpha^2} + \frac{\beta^2 - (\gamma - \alpha)^2 + 4}{4\beta^2} \ge \\ & \frac{\gamma^2 - (\alpha - \beta)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\gamma^2} + \frac{\alpha^2 - (\beta - \gamma)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\alpha^2} + \frac{\beta^2 - (\gamma - \alpha)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\beta^2} = \\ & \frac{2\gamma^2 + 2\alpha\beta}{4\gamma^2} + \frac{2\alpha^2 + 2\beta\gamma}{4\alpha^2} + \frac{2\beta^2 + 2\gamma\alpha}{4\beta^2} = \\ & \frac32 + \frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2}. \end{align*}\]

Therefore it remains to prove that

\[\frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2} \ge \frac32.\].

Which is obviously true by AM-GM.

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Great solution!

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Thank you:-)

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Given constraint: a^2+b^2+c^2+ab+bc+ca\leq2 Now, frac{(ab+1)/(a+b)^2}\geq1/2+1/2{(a+b)(b+c)}/(a+b)^2 Cyclic summing over three variables a,b,c And using A.M-G.M we obtain required result

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This is 2011 USAMO #1 I believe

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Yes it is.

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Comment deleted Jan 29, 2016

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\(a+2b+c<4\) and \(a+b+c < 3\) does not imply that \(b<1\). For example, \(a=0.1, b=1.5, c=0.1\) fulfills the two inequalities but has \(b>1\).

\(a+2b+c<4\) and \(a+b+c>3\), however, does imply \(b<1\).

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I am going to sleep right now. :( Sorry, will continue my solution tomorrow. (It may contain a few minor mistakes!)

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Nice solution, but I have some doubts, I will post my solution too tomorrow

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