$a,b,c,d$ are positive reals with sum $1$. Show that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \ge \frac{1}{2}$ with equality iff $a=b=c=d=\frac{1}{4}$.
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Tags: inequalities, inequalities unsolved
02.03.2013 12:32
$\sum {\frac{a^2}{a+b}} \ge \frac{(\sum{a})^2}{\sum{a+b}} = \frac {a+b+c+d}{2} = \frac{1}{2}$
02.03.2013 16:29
It seems that the following inequality is also true. Let $a$, $b$, $c$ and $d$ are positive numbers such that $a^2+b^2+c^2+d^2=1$. Prove that: \[\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge1\]
13.03.2013 16:16
ACCCGS8 wrote: $a,b,c,d$ are positive reals with sum $1$. Show that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \ge \frac{1}{2}$ with equality iff $a=b=c=d=\frac{1}{4}$. the problem is easy to prove. We have: $\sum \frac{a^2}{a+b}=\sum \left ( a-\frac{ab}{a+b} \right )=\sum a-\sum \frac{ab}{a+b}$ $\geq \sum a-\sum \frac{ab}{2\sqrt{ab}}$ $\geq 1-\frac{1}{2}\sum \sqrt{ab}$ On the other hand, $\sum \sqrt{ab}\leq \sum \frac{a+b}{2}=1$ And the problem is solved. *Another solution: we can use Cauchy- Schwartz.
18.08.2015 18:46
direct application of titu's lemma...(a+b+c+d)^2/2(a+b+c+d)
18.08.2015 18:56
An easy application of a corollary of C-S.
20.10.2016 21:11
Σa^2/(a+b)=Σb^2/(a+b) -- cyclic sums ==> We should prove that cyclic sum (a^2+b^2)/(a+b). But this is obvious: (a^2+b^2)>=1/2(a+b)^2
16.04.2017 15:50
Its Modified Cauchy shawarz
11.02.2018 22:22
aka Titu's Lemma
11.02.2018 22:57
ACCCGS8 wrote: $a,b,c,d$ are positive reals with sum $1$. Show that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \ge \frac{1}{2}$ with equality iff $a=b=c=d=\frac{1}{4}$. Jensen inequality on $f(x)=\frac{1}{1+x}$ \[ \sum \frac{a^2}{a+b}=\sum af(\frac{b}{a}) \geq f(\sum a* \frac{b}{a})=f(\sum a) = f(1) = \frac{1}{2}\]
12.02.2018 18:32
ACCCGS8 wrote: $a,b,c,d$ are positive reals with sum $1$. Show that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \ge \frac{1}{2}$ with equality iff $a=b=c=d=\frac{1}{4}$. Just use $\frac{a^2}{a+b}\ge\frac{3}{4}a-\frac{1}{4}b$ and then add cyclically.
15.02.2020 20:41
Kaskade wrote: ACCCGS8 wrote: $a,b,c,d$ are positive reals with sum $1$. Show that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \ge \frac{1}{2}$ with equality iff $a=b=c=d=\frac{1}{4}$. Just use $\frac{a^2}{a+b}\ge\frac{3}{4}a-\frac{1}{4}b$ and then add cyclically. How did you find it?
16.02.2020 11:14
I think he found it from newton's formula
16.02.2020 11:16
arqady wrote: It seems that the following inequality is also true. Let $a$, $b$, $c$ and $d$ are positive numbers such that $a^2+b^2+c^2+d^2=1$. Prove that: \[\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge1\] a+b+c+d>=2 so that does the work
16.02.2020 14:55
IMO2115 wrote: arqady wrote: It seems that the following inequality is also true. Let $a$, $b$, $c$ and $d$ are positive numbers such that $a^2+b^2+c^2+d^2=1$. Prove that: \[\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge1\] a+b+c+d>=2 so that does the work take $(a,b,c,d)=(1,0,0,0)$. we get: $a+b+c+d=1<2$