If we assume $a+b+c=3$ . We have to prove $\sum \frac{1}{a^2} \geq \frac{63 +7\sum ab -3abc }{27abc}$ After use $ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \geq a^2 + b^2 +c^2$ In LHS the inequality turn to a $3variables$ method. I don`t like such proof !!!

goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) my solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=360828

minhhoang wrote: goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) my solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=360828 Really nice problem I was going to submit this solution but you are faster !! I can`t bealive that this is a new problem . Am I right ? (Where are you arqady ?! )

mahanmath wrote: minhhoang wrote: goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) my solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=360828 Really nice problem I was going to submit this solution but you are faster !! I can`t bealive that this is a new problem . Am I right ? (Where are you arqady ?! ) It's a new problem for me. But this problem is obvious by $uvw$. For the proof it enough to check this inequality for $b=c=1$.

would you please tell us a bit about $UVW$ method? thanks

goodar2006 wrote: would you please tell us a bit about $UVW$ method? thanks You can search (in google or ML , .... ) . Here is a link : UVW

goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) $\Longleftrightarrow \sum_{cyc} \frac{(11a^4+82a^3b+178a^2b^2+54a^2c^2+82ab^3+142abc^2+72ac^3+11b^4+54b^2c^2+72bc^3+7c^4)(a-b)^2}{a^2b^2}$ $\geq 0$ $\Longleftrightarrow \sum_{cyc} \frac{(11a^4+10a^3b+72a^3c+178a^2b^2+54a^2c^2+10ab^3+286abc^2+11b^4+72b^3c+54b^2c^2+7c^4)(a-b)^2}{a^2b^2}$ $\geq 0$ $\Longleftrightarrow \sum_{cyc} \frac{(11a^3c+60a^2b^2+58abc^2+54ac^3+11b^3c+54bc^3+7c^4)(a-b)^2}{ab}$ $\geq 0$ $\Longleftrightarrow \sum_{cyc} \frac{(11a^3c+6a^2b^2+54a^2c^2+112abc^2+11b^3c+54b^2c^2+7c^4)(a-b)^2}{ab}$ $\geq 0$ $\Longleftrightarrow \sum_{cyc} \frac{(11a^3b+6a^2bc+54a^2c^2+11ab^3+6ab^2c+95abc^2+54b^2c^2+18c^4)(a-b)^2}{ab}$ $\geq 0$

goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) LEt $a+b+c=1$ then the inequality will be equivalent to ! $ \left[\frac{ab+bc+ac}{abc}\right]^2 -\frac{2}{abc}+1 \geq \frac{7}{25}(\frac{ab+bc+ac}{abc}+1)^2$ we put $\frac{ab+bc+ac}{abc}=x $ and since $-\frac{2}{abc} \geq -\frac{2x^2}{3}$ so it suffices to prove that : $ \frac{x^2}{3} +1\geq \frac{7}{25}(x+1)^2 \iff (x-9)(2x-3) \geq 0 $ which is true since $x \geq 9$ Done !

minhhoang solution was the best solution but my solution was the Sporovitch solution

goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) Prove my http://diendantoanhoc.net/forum/index.php?showtopic=59440&pid=264854&mode=threaded&start=#entry264854

Since the inequality is homogeneous, we can let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Thus we have to prove \[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} + \frac{1}{(a+b+c)^2} \ge \frac{7}{25}\left(\frac{1}{a+b+c}+1\right)^2\] Let $T = \frac{1}{a+b+c}$. By Cauchy-Schwarz, we have $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge 9 \Longleftrightarrow T \le \frac{1}{9}$, and $3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right) \ge \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2 = 1$, so it remains to prove that \[\frac{1}{3}+T^2 \ge \frac{7}{25}(T+1)^2\] or, after expanding and simplifying, $27T^2 - 21T + 2 \ge 0$. But this factors to $(9T-1)(3T-2) \ge 0$, which is evidently true because $T \le \frac{1}{9}$.

By Cauchy-Schwarz, we have $ \left (1+1+1+\frac{1}{9} \right ) \left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2} \right ) \ge \left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{3(a+b+c)} \right )^2 $ So $ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2} \ge \frac{9}{28}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{3(a+b+c)} )^2 $ $ =\frac{7}{25}( \frac{15}{14a}+\frac{15}{14b}+\frac{15}{14c}+\frac{5}{14(a+b+c)} )^2 $ $ \ge \frac{7}{25}( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c} )^2 $ Because: \[ \frac{1}{15a}+\frac{1}{15b}+\frac{1}{15c} \ge \frac{9}{15a+15b+15c} \]

for #1 we have \[LHS-RHS=\sum\frac{2(a+b)(a-b)^2}{25abc(a+b+c)^2}+\sum\frac{9(a-b)^2}{25a^2b^2}\]

It's easy by Muirhead's inequality. After multiplication we get T(3,2,0)>=T(2,2,1)

The following inequality is also true. Let $a, b, c>0$. Prove that \[\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{(a+b+c)^3}\ge\frac{41}{500}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^3\]

goodar2006 wrote: $a,b,c$ are positive real numbers. prove the following inequality: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{(a+b+c)^2}\ge \frac{7}{25}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c})^2$ (20 points) Generalization Let $a_1,a_2,\cdots,a_n$ be positive real numbers . Prove that$$\sum\limits_{i = 1}^{n}\frac{1}{a^2_i}+\frac{1}{(\sum\limits_{i = 1}^{n}a_i)^2}\ge\frac{n^3+1}{(n^2+1)^2}\left(\sum\limits_{i = 1}^{n}\frac{1}{a_i}+\frac{1}{\sum\limits_{i = 1}^{n}a_i}\right)^2 $$

Clearing the denominator, we get $108S(4,2,0)+108S(3,3,0) \geq 42S(4,1,1)+120S(3,2,1)+54S(2,2,2)$ Obviously this holds by Muirhead inequality.

Since the inequality is in homogenized form, let $a+b+c=1,$ by Cauchy-Schwarz Inequality, we have \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geqslant 9.\]So, \begin{align*} \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1&\geqslant \frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)^2 \\ \iff 9\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+9&\geqslant 7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \end{align*}This follows from adding up the inequalities: \[7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\geqslant 7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\]and \[2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+9 \geqslant 7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\]The first one is evident, we shall prove the second one, by AM-GM, we know \[\frac{1}{a^2}+9\geqslant \frac{6}{a} \implies \frac{1}{a^2}\geqslant \frac{6}{a}-9 \]so \begin{align*} 2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+9&\geqslant 12\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-45\\ &=7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-45 \\ &\geqslant 7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \end{align*}Thus, we have proved the desired inequality. $\quad \blacksquare$