Hmm, I guess you meant $\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c$. Oh yes, nice joke. Darij

By Cauchy Schwartz $(\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b})(a+b+c)=(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b})[(b+c)+(c+a)+(a+b)] \geq (a+b+c)^2$ and the result follows

By Chebyshev and then Nesbitt, $LHS\geq \frac {1}{3}\left(\sum a\right)\left(\sum \frac {2a}{b+c}\right)\geq \sum a$

Indeed very easy! I killed this in 10 seconds. $2\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqq 2\cdot \frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}=a+b+c$ Q.E.D.

but i killed this in 1 seconds,kunny

How can you solve it in such a very short time, spider_boy?

Becaus e it's TRIVIAL

I have for this inequality at least five solutions.For example you can use Chebysev,C.B.S,AM-GM,Jensen or as it follows: a^2/b+c=(a(a+b+c)-(ab+ac))/b+c and by calculus and well known inequality of Nesbitt the conclusion follows.

Here comes two other solutions: (1) Without loss of generality, we can assume a \geq b \geq c, since the inequality is symmetric. Then a^2 \geq b^2 \geq c^2 and \frac {1}{b + c} \geq \frac {1}{a + c} \geq \frac {1}{b + c}. Therefore, by the rearrangement principle, \frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \frac{b^2 + c^2}{b+c} + \frac{c^2 + a^2}{c+a} + \frac{a^2 + b^2}{a+b} = = (b + c) + (c + a) + (a + b) - \frac{2bc}{b+c} + \frac{2ca}{c+a} + \frac{2ab}{a+b} \geq a + b + c, since \frac{b + c}{2} \geq \frac{2bc}{b + c} etc. (2) If we multiply a, b, c with a positive number t, the inequality does not change validity. Therefore, we can assume a + b + c = 1. Now, let f(x) = \frac{2x^2}{1 - x}. Via calculus we then see Jensen coming, and f(a) + f(b) + f(c) \geq 3f(\frac{a + b + c}{3}) = 1.

We have : \frac{a^2}{b+c} + \frac{b+c}{4} \ge 2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}} =a or \frac{a^2}{b+c} \ge a - \frac{b+c}{4}. Do : \frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}

ehku wrote: Do : \frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2} Posted many times... once by me. Chebychev and Nesbitt (again) $\sum{\frac{a^n}{b+c}} \ge \frac{1}{3}\sum{a^{n-1}}\sum{\frac{a}{b+c}} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$

Okie! But there is not only one solution ! Prove that \sum \frac{a^n}{pb+qc} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{p+q} a, b, c, p, q >0.

ehku wrote:

${\frac{a^2}{b+c} + \frac{b+c}{4} \ge 2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}} =a\Rightarrow\sum\frac{a^2}{b+c} \ge \sum(a - \frac{b+c}{4})}=\frac{1}{2}\sum a.$ or $\frac{a^2}{b+c} \geq \frac{4a-b-c}{4}\Rightarrow\sum\frac{a^2}{b+c} \geq \sum\frac{4a-b-c}{4}=\frac{1}{2}\sum a.$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3028019

Tangent Line method Let $a+b+c=1, 0< a,b,c< 1$.The inequality is equivalent to $\sum_{cyc}\frac{2a^2}{1-a}\geq 1$. Let $f(x)=\frac{2x^2}{1-x}$.Then $f(x)\geq \frac{5}{2}x-\frac{1}{2}$ $\Leftrightarrow$ $(3x-1)^2\geq 0$ This holds when $0< x< 1$.So $f(a)+f(b)+f(c)\geq \frac{5}{2}-\frac{3}{2}=1$

arccosinus wrote: Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!) EDITED with exponent 2 over c See also here B-II. Inequalities , Po-Shen Loh , 15 June 2004: 8. (88 Friendship Competition) For $a, b, c > 0:$ $$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+b+c}{2}.$$

Let $a, b$, and $c$ be positive real numbers. Prove that$$\frac{a^2}{b+c } + \frac{b^2}{c+a} + \frac{c^2}{a+b+ \frac{\left(a-b \right)^2}{4(a+b)}} \ge \frac{a+b+c}{2} $$

titu's lemma: $\sum_{cyc} \frac{a^2}{b+c} \geq \frac{(a+b+c)^2}{2(a+b+c)} = \frac{a+b+c}{2}$, multiply by 2 and done

I used Cauchy Schwarz, one term being the LHS, and the other being the cyclic sum of 2(a+b).

For $a, b, c > 0:$ $$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+b+c}{2} $$$$\iff$$$$\frac{a }{b+c} + \frac{b }{c+a} + \frac{c }{a+b} \geq \frac{3}{2} $$

arccosinus wrote: Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!) EDITED with exponent 2 over c Does anyone know the source of this stronger one? For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$

BestChoice123 wrote: arccosinus wrote: Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!) EDITED with exponent 2 over c Does anyone know the source of this stronger one? For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$ This one is stronger ( I found it independently in 2018 but it seems that it was posted by Cezar Lupu earlier than that...) $$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a} \geq \sqrt{3(a^2+b^2+c^2)}$$https://artofproblemsolving.com/community/c6h1761159p11513999

bel.jad5 wrote:

This one is stronger ( I found it independently in 2018 but it seems that it was posted by Cezar Lupu earlier than that...) $$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a} \geq \sqrt{3(a^2+b^2+c^2)}$$https://artofproblemsolving.com/community/c6h1761159p11513999 https://artofproblemsolving.com/community/c6h43001p271771 https://artofproblemsolving.com/community/c6h1438484p8166663

BestChoice123 wrote: arccosinus wrote: Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!) EDITED with exponent 2 over c Does anyone know the source of this stronger one? For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$ $$ LHS-RHS=\sum \frac{(a-b)^2(2a^2+2b^2+3(ab+bc+ca))}{2(a+b+c)(b+c)(c+a)} + [(a-b)^2+(b-c)^2+(c-a)^2]\frac{\sqrt{3(a^2+b^2+c^3)}-(a+b+c)}{2(a+b+c) (a+b+c+\sqrt{3(a^2+b^2+c^2)})}\geqq 0$$

BestChoice123 wrote: For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$ $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq\sqrt[5]{81(a^5+b^5+c^5)}$$for positives $a$, $b$ and $c$ a bit of stronger. There is a nice proof.

Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt{2(b^2+c^2)}}{2}$$

Let $a,b,c$ be positive real numbers. Prove thatLet $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{\sqrt{3(a^2+b^2+c^2)}}{2}\geq \frac{a+\sqrt{2(b^2+c^2)}}{2}$$$$a^2+b^2+c^2\geq\frac{1}{3} \left(a+\sqrt{2(b^2+c^2)}\right)^2\iff 2a^2+b^2+c^2\geq 2a\sqrt{2(b^2+c^2)}$$

Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt[n]{2^{n-1}(b^n+c^n)}}{2}$$Where $5\geq n\in N^+.$ #27 Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{a}{2}+2\sqrt[3]{\frac{b^3+c^3}{2}}$$

sqing wrote: Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt{2(b^2+c^2)}}{2}$$

sqing wrote: Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt[n]{2^{n-1}(b^n+c^n)}}{2}$$Where $5\geq n\in N^+.$ #27 For $n=4$ there is a nice solution.

Let $a, b,c>0.$ Prove that $$\frac{a+k}{b+c+2k} + \frac{b+k}{c+a+2k} +\frac{c+k}{a+b+2k} \geq \frac{3}{2} $$$$\frac{a(a+k)}{b+c+k} + \frac{b(b+k)}{c+a+k} +\frac{c(c+k)}{a+b+k} \geq \frac{a+b+c}{2} $$Where $k\geq 0.$

Let $a, b, c > 0$. Prove that \[\frac{a^2+2bc}{b+c}+\frac{b^2+2ca}{c+a}+\frac{c^2+2ab}{a+b}\geq \frac{3}{2}(a+b+c).\]$$\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$https://artofproblemsolving.com/community/c6h1070208p4651461 https://artofproblemsolving.com/community/c6h1438484p15117577

The inequality is homogenous; WLOG $a+b+c=1$ It sufficies to prove for $a,b,c \in (0,1)$ that, \[\frac{2a^2}{1-a} + \frac{2b^2}{1-b} + \frac{2c^2}{1-c} \geq 1\]provided $a+b+c=1$. Consider $f(x)=\frac{2x^2}{1-x}$. Note that $f''(x)=\frac{4}{(1-x)^3}$ so $f(x)$ is convex on this interval ($(0,1)$). Jensen tells us that, \[\frac{2a^2}{1-a} + \frac{2b^2}{1-b} + \frac{2c^2}{1-c} \geq 3f(\frac{1}{3})=1\]and we are done

Wait, what!!? Titu on\[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b}\]gives\[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b}>\frac{4(a+b+c)^2}{4(a+b+c)}=(a+b+c)\]

nathantareep wrote: \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b}>\frac{4(a+b+c)^2}{2(a+b+c)}=2(a+b+c)>a+b+c\] Sorry, no. See #18 for Titu done properly.

Trivial by holder $$(\sum_{cyc}\frac{a^2}{b+c})(\sum_{cyc}b+c) \geq \left(\sum _{cyc}a\right)^2$$$$\implies 2\sum_{cyc}\frac{a^2}{b+c} \geq \sum_{cyc}a$$

oolite wrote: nathantareep wrote: \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b}>\frac{4(a+b+c)^2}{2(a+b+c)}=2(a+b+c)>a+b+c\] Sorry, no. See #18 for Titu done properly. Ohh I see it was a little error,I have corrected, it see now

nathantareep wrote: oolite wrote: nathantareep wrote: \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b}>\frac{4(a+b+c)^2}{2(a+b+c)}=2(a+b+c)>a+b+c\] Sorry, no. See #18 for Titu done properly. Ohh I see it was a little error,I have corrected, it see now Well actually; If we were to use C-S in Engel form on $\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} $ :

Let $a,b,c$ be positive real numbers. Prove that\[\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{\sqrt{3( a^2+b^2+c^2)}}{2}\]Let $a,b,c,d$ be positive real numbers. Prove or disprove $$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{d+a}+ \frac{d^2}{a+b} \geq \frac{a+b+\sqrt{2(c^2+d^2)}}{2}$$

Isn't simple Titu enough? \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} =\frac{a^2}{\frac{b+c}{2}} + \frac{b^2}{\frac{c+a}{2}} + \frac{c^2}{\frac{a+b}{2}}\]It is easy to see that the denominators add up to $a+b+c$. By Titu's Lemma , we have: \[\frac{a^2}{\frac{b+c}{2}} + \frac{b^2}{\frac{c+a}{2}} + \frac{c^2}{\frac{a+b}{2}} \ge \frac{(a+b+c)^2}{a+b+c} = a+b+c\] So, we are done? $\blacksquare$

this problem must be a joke. It's direct by Titu's lemma

wardtnt1234 wrote: this problem must be a joke. It's direct by Titu's lemma

arccosinus wrote: Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!) A joke for a joke ... WLOG $a\le b\le c$. Write $b=a+x,c=a+x+y$ for $x,y\ge 0$. Substituting and clearing denominators (don't try this at home, kids), we need to prove \begin{align*} 12a^2x^2 + 12a^2xy + 12a^2y^2 + 14ax^3+ 21ax^2y + 27axy^2 + 10ay^3&\\ + 4x^4 + 8x^3y + 13x^2y^2 + 9xy^3 + 2y^4 &\ge 0, \end{align*}which is obviously true

Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}\geq \frac{\sqrt{3(a^2+b^2+c^2)}}{2} \geq \frac{a}{2} + \frac{b^2+c^2}{b+c}\geq \frac{a+\sqrt{2(b^2+c^2)}}{2}$$Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2}\geq \frac{a}{2} + \frac{b^2+c^2}{b+c}$$$$\frac{a^{k+1}}{b^k+c^k} + \frac{b^{k+1}}{c^k+a^k} + \frac{c^{k+1}}{a^k+b^k}\geq \frac{a}{2} + \frac{b^2+c^2}{b+c}$$Where $k \in N^+.$