# Inequality $\sum_{cyc} \sqrt{\frac{a}{b+c+d}}>2$

Let $$a,b,c,d>0$$. I want to prove that $$\sum_{cyc} \sqrt{\frac{a}{b+c+d}}>2.$$

I try using Hölder's inequality : $$\left(\sum_{cyc} b+c+d\right)\left(\sum_{cyc} \sqrt{\frac{a}{b+c+d}}\right)^2\geq\left(\sum_{cyc} \sqrt[3] a\right)^3$$

so it would be enough to prove $$(\sqrt[3] a+ \sqrt[3] b+ \sqrt[3] c+ \sqrt[3] d)^3\geq 12(a+b+c+d)$$

• Compare math.stackexchange.com/q/1003117 for a similar problem. Some of the solutions can be generalised from three to four variables. Dec 6, 2019 at 12:18

Your last inequality is wrong, try for example $$a=b=\frac1{100}$$ and $$c=d=\frac{1}{1000}$$. However, we can proceed as follows:

Lemma. For $$a,b,c,d>0$$, we have $$\sqrt{\frac{a}{b+c+d}}\geq \frac{2a}{a+b+c+d}.$$

Proof. Note that $$(a-b-c-d)^2\geq0$$ so that $$(a + b + c + d)^2-4 a (b + c + d)\geq 0$$ which means that $$(a + b + c + d)^2\geq 4a (b + c + d).$$

Hence, we have $$\frac{a}{b+c+d}\geq\frac{4a^2}{(a+b+c+d)^2},$$ which proves the Lemma. $$\square$$

We can now conclude since $$\sum_{\text{cyc}} \sqrt{\frac{a}{b+c+d}}\geq \sum_{\text{cyc}} \frac{2a}{a+b+c+d}=2.$$

Equality occurs only if $$a=b=c=d=0$$ which is not possible.

• The Lemma also follows easily from the AM-GM inequality, in fact it is a generalisation of the first solution in math.stackexchange.com/a/2237066 from three to four variables. Dec 6, 2019 at 12:16
• @MartinR Thank you for noting that! Dec 6, 2019 at 12:57