$\sum {\frac{a^2}{a+b}} \ge \frac{(\sum{a})^2}{\sum{a+b}} = \frac {a+b+c+d}{2} = \frac{1}{2}$

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: \[\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge1\]

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.

direct application of titu's lemma...(a+b+c+d)^2/2(a+b+c+d)

An easy application of a corollary of C-S.

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

Its Modified Cauchy shawarz

aka Titu's Lemma

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 $f(x)=\frac{1}{1+x}$ \[ \sum \frac{a^2}{a+b}=\sum af(\frac{b}{a}) \geq f(\sum a* \frac{b}{a})=f(\sum a) = f(1) = \frac{1}{2}\]

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.

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?

I think he found it from newton's formula

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

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: $a+b+c+d=1<2$