Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}.\]
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Tags: inequalities, function, inequalities unsolved
10.04.2011 18:41
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.
11.04.2011 19:53
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.
28.06.2011 01:00
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.
22.10.2011 13:47
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
28.03.2014 03:01
Stronger inequality: Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality \[\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq \frac{3}{\sqrt{2}}.\]
28.03.2014 18:44
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 $c\rightarrow0^+$.
29.03.2014 04:09
Thank arqady. Let $a,b,c$ be positive reals such that $a+b+c=3$. Prove the inequality\[\frac{a}{\sqrt{a}+b}+\frac{b}{\sqrt{b}+c}+\frac{c}{\sqrt{c}+a}\geq \frac{3}{2}.\]
29.03.2014 04:26
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.
30.03.2014 08:19
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
02.06.2014 10:17
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.
27.01.2015 20:14
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} $ +$c\sqrt{b}$ +$a\sqrt{c}$ $<=$ ($a$+$b$+$c$+$ab$+$ac$+$bc$)$/2$<=$3$
18.07.2015 11:45
http://www.artofproblemsolving.com/community/c6h1116140p5109299: Let $a,b,c$ be nonnegative reals such that $a+b+c=3$. Prove the inequality$$\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\geq 1.$$
18.02.2016 10:13
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 \[\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\ge \frac{3}{2}\]
24.05.2016 08:23
http://wenku.baidu.com/view/dec4beb46137ee06eef91838.html
24.05.2016 08:38
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.
07.06.2016 11:33
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.
07.07.2016 14:39
EasternMath how?
07.07.2016 14:40
And how exactly does Jensen inequality go without functions and such things?(In algebra only)
25.03.2018 01:03
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.
26.03.2018 16:28
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.
27.03.2018 01:01
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!!
27.03.2018 02:07
27.03.2018 05:42
GeronimoStilton wrote:
It should be, \[\geq \frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\]
22.04.2023 21:17
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$$3$
06.05.2023 00:04
Let $a^2=x, b^2=y\text{, and }c^2=z$ Thus the conditions become $\sum_{cyc}a=\sum_{cyc}\sqrt{x}=3$ Furthermore $\sum_{cyc}\frac{a^2}{a+b^2}=\sum_{cyc}\frac{x}{\sqrt{x}+y}\overset{\text{C-S}}{\ge}\frac{(\sum_{cyc}x)^2}{\sum_{cyc}\sqrt{x}+\sum_{cyc}x}$ Thus the inequalities is equivalent to $\frac{(\sum_{cyc}a^2)^2}{3+\sum_{cyc}a^2}\ge\frac{3}{2}\Longrightarrow \bigg(\sum_{cyc}a^2\bigg)\ge3(6-\sum_{cyc}ab)$ Thus the inequality is equivalent to proving the 2 following inequalities and multiplying them together. $\sum_{cyc}a^2\ge6-\sum_{cyc}ab\Longleftrightarrow \bigg(\sum_{cyc}a\bigg)^2\ge3\sum_{cyc}ab$ $\square$ $\sum_{cyc}a^2\ge3\Longleftrightarrow \sum_{cyc}a^2\ge\sum_{cyc}ab$ $\square$ $QED$
06.05.2023 05:03
Let $a,b,c\ge 0$ and $\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{3}{2}$. Prove that $$a+b+c\geq \frac{3}{2}$$
06.05.2023 07:28
Vrangr wrote: GeronimoStilton wrote:
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
06.05.2023 14:48
Seungjun_Lee wrote: Vrangr wrote: GeronimoStilton wrote:
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.
06.05.2023 18:11
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