LOL http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=401290 By the way, Inequalities have their own forum outside Algebra due to their mass popularity.

I am aware of the inequalities subforum (it is huge enough so you can hardly miss it) Although the competition rules(for CMO) state that problems shall be 1. Algebra 2. Combinatorics 3. Geometry 4. Number theory Because of that I decided exceptionally to put this inequality in this subforum, I was not aware it had been posted before.

Notice that $\frac{a^2}{a+b^2}=a-\frac{ab^2}{a+b^2}$. Hence, it is enough to prove that $ \frac{ab^2}{a+b^2} +\frac{bc^2}{b+c^2} +\frac{ca^2}{c+a^2} \leqslant \frac{3}{2}$. By AM-GM we have that $a+b^2 \geqslant 2b\sqrt{a}$, and other similar inequalities. Hence $\frac{ab^2}{a+b^2}\leqslant \frac{ab^2}{2b\sqrt{a}}=\frac{b\sqrt{a}}{2}$. It is enough to prove that $\sqrt{a} b +\sqrt{b}c +\sqrt{c} a \leqslant 3$. Since $a+b+c=3$, and by the concavity of the function $f(x)=\sqrt{x}$, using Jensen's inequality we have: $\sqrt{a}b +\sqrt{b}c +\sqrt{c} a \leqslant 3\sqrt{\frac{ab +bc+ca}{3}}\leqslant 3$, as desired.

it suffices to prove $\sum_{cyc}\frac{ab^2}{a+b^2}\le\frac{3}{2}$ $LHS\le\sum_{cyc}\frac{aB^2}{2b\sqrt{a}}=\frac{1}{2}\sum_{cyc}\sqrt{a}b \le\frac{1}{2}\sum_{cyc}\frac{ab+a}{2}\le\frac{3}{2}$ QED

Stronger inequality: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq \frac{3}{\sqrt{2}}.\]

sqing wrote: Stronger inequality: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq \frac{3}{\sqrt{2}}.\] It's wrong! Try $a=b=1.5$ and $c\rightarrow0^+$.

Thank arqady. Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality\[\frac{a}{\sqrt{a}+b}+\frac{b}{\sqrt{b}+c}+\frac{c}{\sqrt{c}+a}\geq \frac{3}{2}.\]

sqing wrote: Thank arqady. Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality\[\frac{a}{\sqrt{a}+b}+\frac{b}{\sqrt{b}+c}+\frac{c}{\sqrt{c}+a}\geq \frac{3}{2}.\] The inequality is equivalent with $x^2+y^2+z^2=3$ and \[ \frac{x^2}{x+y^2}+\frac{y^2}{y+z^2}+\frac{z^2}{z+x^2} \geq \frac{3}{2}\] The proof is similar to enndb0x.

let $a,b,c\geq0$ and $a+b+c=1$ prove that\[\dfrac{a}{\sqrt{a+b^2}}+\dfrac{b}{\sqrt{b+c^2}}+\dfrac{c}{\sqrt{c+a^2}} \leq \dfrac{3}{2}.\] here

Matematika wrote: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}.\] The inequality is equivalent to \[ \frac{a^2}{\frac{1}{3} a (a+b+c)+b^2}+\frac{c^2}{a^2+\frac{1}{3} c (a+b+c)}+\frac{b^2}{\frac{1}{3} b (a+b+c)+c^2}-\frac{3}{2} \geq 0 \]. After bringing everything to one fraction we observe that it is enough to prove that $f(a,b,c) \geq 0$ where \[ f(a,b,c)=3 a^5 b+6 a^4 b^2+a^4 b c+9 a^4 c^2-6 a^3 b^3+3 a^3 b^2 c-9 a^3 b c^2-6 a^3 c^3+9 a^2 b^4-9 a^2 b^3 c-21 a^2 b^2 c^2+3 a^2 b c^3+6 a^2 c^4+a b^4 c+3 a b^3 c^2-9 a b^2 c^3+a b c^4+3 a c^5+3 b^5 c+6 b^4 c^2-6 b^3 c^3+9 b^2 c^4 \]. Inequality is cyclic and we may assume that $c$ is smallest among $a,b,c$. If $a\geq b$ then $a=c+x+y$ and $b=c+x$ for some $x,y \geq 0$ and \[ f(a,b,c)=f(c+x+y,c+x,c) = 60 c^4 x^2+60 c^4 x y+60 c^4 y^2+164 c^3 x^3+255 c^3 x^2 y+243 c^3 x y^2+76 c^3 y^3+163 c^2 x^4+344 c^2 x^3 y+366 c^2 x^2 y^2+185 c^2 x y^3+31 c^2 y^4+71 c x^5+191 c x^4 y+234 c x^3 y^2+151 c x^2 y^3+43 c x y^4+3 c y^5+12 x^6+39 x^5 y+57 x^4 y^2+48 x^3 y^3+21 x^2 y^4+3 x y^5 \geq 0 \]. If $b\geq a$ then $b=c+x+y$ and $a=c+x$ for some $x,y \geq 0$ and \[ f(a,b,c)=f(c+x,c+x+y,c) = 60 c^4 x^2+60 c^4 x y+60 c^4 y^2+164 c^3 x^3+237 c^3 x^2 y+225 c^3 x y^2+76 c^3 y^3+163 c^2 x^4+308 c^2 x^3 y+312 c^2 x^2 y^2+167 c^2 x y^3+31 c^2 y^4+71 c x^5+164 c x^4 y+180 c x^3 y^2+115 c x^2 y^3+34 c x y^4+3 c y^5+12 x^6+33 x^5 y+42 x^4 y^2+30 x^3 y^3+9 x^2 y^4 \geq 0 \] Done.

enndb0x wrote: Notice that $\frac{a^2}{a+b^2}=a-\frac{ab^2}{a+b^2}$. Hence, it is enough to prove that $ \frac{ab^2}{a+b^2} +\frac{bc^2}{b+c^2} +\frac{ca^2}{c+a^2} \leqslant \frac{3}{2}$. By AM-GM we have that $a+b^2 \geqslant 2b\sqrt{a}$, and other similar inequalities. Hence $\frac{ab^2}{a+b^2}\leqslant \frac{ab^2}{2b\sqrt{a}}=\frac{b\sqrt{a}}{2}$. It is enough to prove that $\sqrt{a} b +\sqrt{b}c +\sqrt{c} a \leqslant 3$. Since $a+b+c=3$, and by the concavity of the function $f(x)=\sqrt{x}$, using Jensen's inequality we have: $\sqrt{a}b +\sqrt{b}c +\sqrt{c} a \leqslant 3\sqrt{\frac{ab +bc+ca}{3}}\leqslant 3$, as desired. with out using jensen's inequality we can solve it too. $b\sqrt{a} $ +$c\sqrt{b}$ +$a\sqrt{c}$ $<=$ ($a$+$b$+$c$+$ab$+$ac$+$bc$)$/2$<=$3$

http://www.artofproblemsolving.com/community/c6h1116140p5109299: Let $a,b,c$ be nonnegative reals such that $a+b+c=3$. Prove the inequality$$\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\geq 1.$$

Matematika wrote: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}.\] Ukrainian National Olympiad 2016 http://www.artofproblemsolving.com/community/c6h424699p2403254 Let $a,b,c>0,a+b+c=3$,Prove that \[\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\ge \frac{3}{2}\]

http://wenku.baidu.com/view/dec4beb46137ee06eef91838.html

here is a article

Attachments:
lol.pdf (65kb)

I see a good application of Titu's lemma. Apply this, yielding 9/(3+a^2+b^2+c^2)<=the expression. Now apply am-gm to get a^2+b^2+c^2>=3(abc)^(2/3)=3 at equality. Plugging this in yield the expression>=9/(3+3)=3/2.

FCBarcelona wrote: enndb0x wrote: Notice that $\frac{a^2}{a+b^2}=a-\frac{ab^2}{a+b^2}$. Hence, it is enough to prove that $ \frac{ab^2}{a+b^2} +\frac{bc^2}{b+c^2} +\frac{ca^2}{c+a^2} \leqslant \frac{3}{2}$. By AM-GM we have that $a+b^2 \geqslant 2b\sqrt{a}$, and other similar inequalities. Hence $\frac{ab^2}{a+b^2}\leqslant \frac{ab^2}{2b\sqrt{a}}=\frac{b\sqrt{a}}{2}$. It is enough to prove that $\sqrt{a} b +\sqrt{b}c +\sqrt{c} a \leqslant 3$. Since $a+b+c=3$, and by the concavity of the function $f(x)=\sqrt{x}$, using Jensen's inequality we have: $\sqrt{a}b +\sqrt{b}c +\sqrt{c} a \leqslant 3\sqrt{\frac{ab +bc+ca}{3}}\leqslant 3$, as desired. with out using jensen's inequality we can solve it too. $b\sqrt{a} $ +$c\sqrt{b}$ +$a\sqrt{c}$ $<=$ ($a$+$b$+$c$+$ab$+$ac$+$bc$)$/2$<=$3$ Cauchy inequality can also solve the last step.

EasternMath how?

And how exactly does Jensen inequality go without functions and such things?(In algebra only)

Another, similar approach: The given inequality is equivalent to $ \frac{ab^2}{a+b^2} +\frac{bc^2}{b+c^2} +\frac{ca^2}{c+a^2} \leqslant \frac{3}{2}$ . Using Cauchy-Schwarz's inequality one obtains: $(a+b^2)\geqslant \frac{4ab}{(a+1)}$. Following this, we have: $ \frac{ab^2}{a+b^2} +\frac{bc^2}{b+c^2} +\frac{ca^2}{c+a^2} \leqslant \sum_{cyc}\frac{ab^2(a+1)}{4ab} = \sum_{cyc}\frac{b(a+1)}{4} = \sum_{cyc}\frac{ab+b}{4} = \frac{(\sum_{cyc}{ab})+3}{4}$. Since $\sum_{cyc}{ab} \leqslant 3$, we obtatin the desired result.

Matematika wrote: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}.\] The inequality is equivalent to $\sum_{cyc}\frac{3a^2}{a(a+b+c)+3b^2}\geq \frac{3}{2}$ Clearing the denominator, we get $3\sum_{cyc}a^5b+6\sum_{cyc}a^4b^2+9\sum_{cyc}a^4c^2+\sum_{cyc}a^4bc+3\sum_{cyc}a^3b^2c\geq 6\sum_{cyc}a^3b^3+9\sum_{cyc}a^3bc^2+21a^2b^2c^2$ By AM-GM inequality, $3\sum_{cyc}a^4b^2+3\sum_{cyc}a^4c^2\geq 6\sum_{cyc}a^3b^3$ and $\sum_{cyc}a^5b+\sum_{cyc}a^4bc\geq 2\sum_{cyc}a^3bc^2$ and $2\sum_{cyc}a^5b\geq 2\sum_{cyc}a^3bc^2$ and $3\sum_{cyc}a^4b^2\geq 3\sum_{cyc}a^3bc^2$ and $2\sum_{cyc}a^4c^2\geq 2\sum_{cyc}a^3bc^2$ and $4\sum_{cyc}a^4c^2+3\sum_{cyc}a^3b^2c\geq 21a^2b^2c^2$ So we're done.

Rearrange the numerator to cancel out; you obtain a simpler inequality equivalent to the initial ! Note: I dont know why, but the trick-ish where the numerator of the terms of an inequality is decomposed in terms of the denominator so they divide and obtain a simpler inequality; I've seen this prominently in Balkan regions problems (BMO, Bosnia etc.. ) I am curious as to why (I honestly love this trick tbh), I might be mistaken so correct me if so!!

Titu's Lemma

GeronimoStilton wrote:

Titu's Lemma

It should be, \[\geq \frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\]

Matematika wrote: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}.\] We use $AM$ $\ge$ $GM$ after let's show that $\sum_{cyc}{\frac{a^2}{1+a^2+2b^2}}$ $\ge$ $\frac{3}{4}$ let $a^2=x$ $b^2=y$$c^2=z$ by TİTU lemma $\sum_{cyc}{\frac{x^2}{2xy+x^2+x}}$$\ge$ $\frac{(x+y+z)^2}{x+y+z+3(xy+xz+yz)}$$\ge$ $\frac{x+y+z}{x+y+z+1}$$\ge$$\frac{3}{4}$ $\Longrightarrow$ $x+y+z$$\ge$$3$ $a^2+b^2+c^2$$\ge$$\frac{1}{3}$$×$$(a+b+c)^2$$\ge$$3$

Let $a^2=x, b^2=y\text{, and }c^2=z$ Thus the conditions become $\sum_{cyc}a=\sum_{cyc}\sqrt{x}=3$ Furthermore $\sum_{cyc}\frac{a^2}{a+b^2}=\sum_{cyc}\frac{x}{\sqrt{x}+y}\overset{\text{C-S}}{\ge}\frac{(\sum_{cyc}x)^2}{\sum_{cyc}\sqrt{x}+\sum_{cyc}x}$ Thus the inequalities is equivalent to $\frac{(\sum_{cyc}a^2)^2}{3+\sum_{cyc}a^2}\ge\frac{3}{2}\Longrightarrow \bigg(\sum_{cyc}a^2\bigg)\ge3(6-\sum_{cyc}ab)$ Thus the inequality is equivalent to proving the 2 following inequalities and multiplying them together. $\sum_{cyc}a^2\ge6-\sum_{cyc}ab\Longleftrightarrow \bigg(\sum_{cyc}a\bigg)^2\ge3\sum_{cyc}ab$ $\square$ $\sum_{cyc}a^2\ge3\Longleftrightarrow \sum_{cyc}a^2\ge\sum_{cyc}ab$ $\square$ $QED$

Let $a,b,c\ge 0$ and $\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}$. Prove that $$a+b+c\geq \frac{3}{2}$$

Vrangr wrote: GeronimoStilton wrote:

Titu's Lemma

It should be, \[\geq \frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\] $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$ Therefore $a^2 +b^2 +c^2 \ge a+b+c$ Hence the result is correct

Seungjun_Lee wrote: Vrangr wrote: GeronimoStilton wrote:

Titu's Lemma

It should be, \[\geq \frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\] $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$ Therefore $a^2 +b^2 +c^2 \ge a+b+c$ Hence the result is correct This reasoning is false. Since $\sum_{cyc}a^2\ge\sum_{cyc}a$ you conclude that: $\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\ge \frac{(a+b+c)^2}{2a + 2b + 2c}$ which is false. It is false because if we cancel the $(a+b+c)^2$ we are left with: $\frac{1}{\sum_{cyc}a+\sum_{cyc}a^2}\ge\frac{1}{2\sum_{cyc}a}\Longrightarrow 2\sum_{cyc}a\ge\sum_{cyc}a+\sum_{cyc}a^2\Longleftrightarrow \sum_{cyc}a\ge\sum_{cyc}a^2$ which contradicts the statement Thus it should be $\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\le \frac{(a+b+c)^2}{2a + 2b + 2c}$ You could also prove it is false from the fact that when you divide by a smaller number you get a bigger number. Since $(a+b+c)^2+a^2+b^2+c^2\ge 2(a+b+c)$ thus you divide by a smaller number which implies that the result is larger that the original.

F10tothepowerof34 wrote: This reasoning is false. Since $\sum_{cyc}a^2\ge\sum_{cyc}a$ you conclude that: $\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\ge \frac{(a+b+c)^2}{2a + 2b + 2c}$ which is false. It is false because if we cancel the $(a+b+c)^2$ we are left with: $\frac{1}{\sum_{cyc}a+\sum_{cyc}a^2}\ge\frac{1}{2\sum_{cyc}a}\Longrightarrow 2\sum_{cyc}a\ge\sum_{cyc}a+\sum_{cyc}a^2\Longleftrightarrow \sum_{cyc}a\ge\sum_{cyc}a^2$ which contradicts the statement Thus it should be $\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\le \frac{(a+b+c)^2}{2a + 2b + 2c}$ You could also prove it is false from the fact that when you divide by a smaller number you get a bigger number. Since $(a+b+c)^2+a^2+b^2+c^2\ge 2(a+b+c)$ thus you divide by a smaller number which implies that the result is larger that the original. Thank you very much I made a mistake typing with a phone outside.. My reasoning was incorrect