Prove that $\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{b+c+2}}+\frac{1}{\sqrt{c+d+2}}+\frac{1}{\sqrt{d+a+2}}\le 2$

Question

Let $a,b,c,d\in \mathbb{R^+}$ such that $abcd=1$. Prove that $$\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{b+c+2}}+\frac{1}{\sqrt{c+d+2}}+\frac{1}{\sqrt{d+a+2}}\le 2$$

---

By Cauchy-Schwarz:

$$\text{LHS}^2=\left(\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{b+c+2}}+\frac{1}{\sqrt{c+d+2}}+\frac{1}{\sqrt{d+a+2}}\right)^2$$

$$\le 4\left(\sum \frac{1}{a+b+2}\right)\le 4\left(\sum \frac{1}{4}\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\right)$$

And need to prove $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}\le 2$$ for $abcd=1$

I tried AM-GM and Vasile inequality but failed

Answer 1

As you've shown we can use Cauchy-Schwarz, or AM-GM:

$$\sum \frac{1}{\sqrt{a+b+2}} \le \sum\left(\frac{1}{4}+\frac{1}{a+b+2}\right)$$

So it is enough to prove:

$$\sum \frac{1}{a+b+2} \leq 1$$

Notice that using AM-GM, we have:

$$ \begin{aligned} \frac{1}{a+b+2}+\frac{1}{c+d+2} &\leq \frac{1}{2\sqrt{ab}+2}+\frac{1}{2\sqrt{cd}+2}\\ &=\frac{1}{2}\left(\frac{1}{\sqrt{ab}+1}+\frac{\sqrt{ab}}{1+\sqrt{ab}}\right)\\ &=\frac{1}{2} \end{aligned} $$

Summing with other similar inequality, the proof is completed.

Answer 2

By Jensen $$\sum_{cyc}\frac{1}{\sqrt{a+b+2}}\leq4\sqrt{\frac{\sum\limits_{cyc}\frac{1}{a+b+2}}{4}}.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{1}{a+b+2}\leq1.$$ Indeed, let $a=\frac{x^3}{yzt}$, $b=\frac{y^3}{xzt}$, $c=\frac{z^3}{xyt}$ and $d=\frac{t^3}{xyz}$, where $x$, $y$, $z$ and $t$ are positive numbers.

Thus, $$\sum\limits_{cyc}\frac{1}{a+b+2}=\sum_{cyc}\frac{1}{\frac{x^3}{yzt}+\frac{y^3}{xzt}+2}=\sum_{cyc}\frac{xyzt}{x^4+y^4+2xyzt}\leq$$ $$\leq\sum_{cyc}\frac{xyzt}{xy(x^2+y^2)+2xyzt}=\sum_{cyc}\frac{zt}{x^2+y^2+2zt}$$ and it's enough to prove that: $$\sum_{cyc}\frac{zt}{x^2+y^2+2zt}\leq1$$ or $$\sum_{cyc}\left(\frac{zt}{x^2+y^2+2zt}-\frac{1}{2}\right)\leq-1$$ or $$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+2zt}\geq2,$$ which is true by AM-GM: $$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+2zt}\geq\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+z^2+t^2}=2.$$ Done!

Answer 3

Another way.

We'll prove that: $$\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{c+d+2}}\leq1.$$ Indeed, we need to prove that $$\left(\sqrt{a+b+2}+\sqrt{c+d+2}\right)^2\leq(a+b+2)(c+d+2)$$ or $$2\sqrt{(a+b+2)(c+d+2)}\leq(a+b)(c+d)+a+b+c+d,$$ which is true by AM-GM twice: $$(a+b)(c+d)+a+b+c+d\geq$$ $$\geq4+a+b+c+d=a+b+2+c+d+2\geq2\sqrt{(a+b+2)(c+d+2)}.$$ Similarly, $$\frac{1}{\sqrt{b+c+2}}+\frac{1}{\sqrt{d+a+2}}\leq1$$ and we are done!