https://www.dropbox.com/s/4h5688xnbmhlris/Problema%203.pdf

How about the condition no two are equal?

I have somewhat a different solution. Let $\frac{a}{b}=u,\frac{b}{c}=v$. Then $\frac{c}{d}=\frac{1}{u},\frac{d}{a}=\frac{1}{v}$. Then we have $u+v+\frac{1}{u}+\frac{1}{v}=4$, and we have to maximise $\mathcal{P}=uv+\frac{u}{v}+\frac{v}{u}+\frac{1}{uv}$. Now see $u,v$ both cannot be negative, and if both are positive then $u+\frac{1}{u}\ge 2,v+\frac{1}{v}\ge 2\implies u+v+\frac{1}{u}+\frac{1}{v}\ge 4$. thus equality holds and $u=v=1$ contradicting the distinctness. WLOG $u>0,v<0$. Then $v+\frac{1}{v}\le -2$, and $u+\frac{1}{u}=4-\left(v+\frac{1}{v}\right)\ge 6$. But we have \[ \mathcal{P}=\left(u+\frac{1}{u}\right)\left(v+\frac{1}{v}\right)\le -12 \] and considering equality we have $u+\frac{1}{u}=6,v+\frac{1}{v}=-2$ and see $\boxed{(a,b,c,d)=(3+2\sqrt{2},1,-1,-3+2\sqrt{2})}$ work. I have a different answer than the one above, hope I am correct.

You are correct. Alternative would be by homogenizing you can say $ac=bd=1$ or $ac=bd=-1$. The first case is impossible when $a,b,c,d$ have to be different and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$ In the second case we get $(a+c)(b+d)=-4$ and we want to minimize $a^2+c^2+b^2+d^2=(a+c)^2+(b+d)^2+4 \le 12$. Dividing by $ac=-1$ gives the original thing is maximized to $-12$.

Let $\frac{a}{b}=x,\frac{b}{c}=y,y=kx(k<0),t=k+\frac{1}{k}+2<0$, hence $(k+1)(x+\frac{1}{kx})=4$, $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=k+\frac{1}{k}+k(x^2+\frac{1}{k^2x^2})$ $=k+\frac{1}{k}+k(\frac{16}{(k+1)^2}-\frac{2}{k})=t+\frac{16}{t}-4\leq -12$. when $t=-4$,$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=-12$.

inequality problem with cyclic independents Israel 2013, stage 1, problem 10 Pan African Math Olympiad 2018, Problem 5

Let $a$ and $b$ are positive real numbers. Prove that$$\frac{(a-b)^2}{2(a+b)}\le \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\le\frac{(a-b)^2}{\sqrt{2}(a+b)}$$(The problems proposed to the Central American Olympiad 2015)

sqing wrote: Let $a$ and $b$ are positive real numbers. Prove that$$\frac{(a-b)^2}{2(a+b)}\le \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\le\frac{(a-b)^2}{\sqrt{2}(a+b)}$$(The problems proposed to the Central American Olympiad 2015)

Attachments:

sqing wrote: Let $a$ and $b$ are positive real numbers. Prove that$$\frac{(a-b)^2}{2(a+b)}\le \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\le\frac{(a-b)^2}{\sqrt{2}(a+b)}$$(The problems proposed to the Central American Olympiad 2015) We have $\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}=\frac{\frac{a^2+b^2}{2}-ab}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}=\frac{(a-b)^2}{\sqrt{2}(\sqrt{a^2+b^2}+\sqrt{2ab})}$ Hence, it's left to prove $$a+b \leq \sqrt{a^2+b^2}+\sqrt{2ab} \leq \sqrt{2}(a+b)$$Let $\sqrt{a^2+b^2}=x, \sqrt{2ab}=y$, the inequality becomes $$x^2+y^2 \leq (x+y)^2 \leq 2(x^2+y^2)$$$$0 \leq 2xy \leq x^2+y^2$$This is obviously true.

nmtuan2001 wrote: sqing wrote: Let $a$ and $b$ are positive real numbers. Prove that$$\frac{(a-b)^2}{2(a+b)}\le \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\le\frac{(a-b)^2}{\sqrt{2}(a+b)}$$(The problems proposed to the Central American Olympiad 2015) We have $\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}=\frac{\frac{a^2+b^2}{2}-ab}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}=\frac{(a-b)^2}{\sqrt{2}(\sqrt{a^2+b^2}+\sqrt{2ab})}$ Hence, it's left to prove $$a+b \leq \sqrt{a^2+b^2}+\sqrt{2ab} \leq \sqrt{2}(a+b)$$Let $\sqrt{a^2+b^2}=x, \sqrt{2ab}=y$, the inequality becomes $$x^2+y^2 \leq (x+y)^2 \leq 2(x^2+y^2)$$$$0 \leq 2xy \leq x^2+y^2$$This is obviously true. Nice. Thanks.

leonardg wrote: sqing wrote: Let $a$ and $b$ are positive real numbers. Prove that$$\frac{(a-b)^2}{2(a+b)}\le \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\le\frac{(a-b)^2}{\sqrt{2}(a+b)}$$(The problems proposed to the Central American Olympiad 2015) $$\frac{(a-b)^2}{2(a+b)}\le \sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\le\frac{(a-b)^2}{\sqrt{2}(a+b)} \Leftrightarrow$$$$(\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab})\frac{(a-b)^2}{2(a+b)}\le (\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab}) [\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}]\le(\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab}) \frac{(a-b)^2}{\sqrt{2}(a+b)} \Leftrightarrow$$$$(\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab})\frac{(a-b)^2}{2(a+b)}\le \dfrac {(a-b)^2} {2}\le(\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab}) \frac{(a-b)^2}{\sqrt{2}(a+b)} $$Using Cauchy-Shwarz inequality $$ \sqrt{\dfrac {a^2+b^2} {2}} +\sqrt{ab} \le a+b$$hence we have $$(\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab})\frac{(a-b)^2}{2(a+b)}\le \dfrac {(a-b)^2} {2}$$also we have $$ \sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab}\ge \sqrt {\dfrac {a^2+b^2} {2}+ab}=\dfrac {a+b} {\sqrt{2}}$$so we have $$\dfrac {(a-b)^2} {2}\le(\sqrt{\dfrac{a^2+b^2} {2}}+\sqrt{ab}) \frac{(a-b)^2}{\sqrt{2}(a+b)} $$