Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $\prod\limits_{cyc}(a+b)\neq0$. Prove that: $$\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}+\frac{d-a}{\sqrt{a+b}}\geq0$$

The equality occurs also for $a=c$ and $b=d$.

My trying.

We need to prove that $$\sum_{cyc}\frac{a+c-b-c}{\sqrt{b+c}}\geq0$$ or $$(a+c)\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{d+a}}\right)+(b+d)\left(\frac{1}{\sqrt{c+d}}+\frac{1}{\sqrt{a+b}}\right)\geq\sum_{cyc}\sqrt{a+b}$$ and what is the rest?

Thank you!

  • $\begingroup$ Let $t=\sum_{cyc}a, x=a+b, y=a+c, z=a+d$ then your last inequality is $y(1/\sqrt z+1/\sqrt {t-z})+(t-y)(1/\sqrt x+1/\sqrt {t-x}) \ge \sqrt x+\sqrt {t-x}+\sqrt z+\sqrt {y-z}$ $\endgroup$ – didgogns Apr 1 '17 at 12:44

Since the sign of $\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}+\frac{d-a}{\sqrt{a+b}}$ is preserved when $(a,b,c,d)$ is replaced with $(ax,bx,cx,dx)$ for any positive $x$, we can assume that $a+b+c+d=1$ WLOG.

Now let $x=a+b,y=a+c,z=a+d$ and from OP's last inequality, we get $$y\left(\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}}\right)+(1-y)\left(\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}}\right) \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$

To prove this, let's consider this inequality.$$\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$ And $\sqrt z+\sqrt {1-z}$ is maximized at $\sqrt 2$ when $z=1/2$. $$\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt 2$$ We can verify this by using calculus, graphing tools, etc.

Therefore, $\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$ and similarly $$\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$ Now we get $$\begin{align}&y\left(\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}}\right)+(1-y)\left(\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}}\right) \\\ge& y(\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z})+(1-y)(\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z})\\=&\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}\end{align}$$


  • $\begingroup$ Well I proved $\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt 2$ with calculus and thought this proof is cleaner so edited my answer, but it's wrong... $\endgroup$ – didgogns Apr 2 '17 at 15:26

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