Given positive real numbers $a$, $b$, $c$, $d$, $e$ with $\sum_{\text{cyc}}\,\frac{1}{4+a}=1$, prove that $\sum_{\text{cyc}}\,\frac{a}{4+a^2}\le1$.

Question

Let $a, b, c, d, e$ be positive real numbers such that $$\dfrac{1}{4+a} + \dfrac{1}{4+b} +\dfrac{1}{4+c} +\dfrac{1}{4+d} +\dfrac{1}{4+e} = 1.$$ Prove that $$\dfrac{a}{4+a^{2}} + \dfrac{b}{4+b^{2}} +\dfrac{c}{4+c^{2}} +\dfrac{d}{4+d^{2}} +\dfrac{e}{4+e^{2}} \leq 1.$$

My question is how to prove this inequality by using AM-GM inequality?

My solution (using the Chebyshev inequality).

Since $\dfrac{1}{4+a}+\dfrac{1}{4+b}+\dfrac{1}{4+c}+\dfrac{1}{4+d}+\dfrac{1}{4+e} =1,$ we have

$$1 = \dfrac{1}{4+a}+\dfrac{1}{4+b}+\dfrac{1}{4+c}+\dfrac{1}{4+d}+\dfrac{1}{4+e} \geq \dfrac{a}{4+a^2}+\dfrac{b}{4+b^2}+\dfrac{c}{4+c^2}+\dfrac{d}{4+d^2}+\dfrac{e}{4+e^2}$$ $$\Leftrightarrow \dfrac{1-a}{(4+a)(4+a^2)}+\dfrac{1-b}{(4+b)(4+b^2)}+\dfrac{1-c}{(4+c)(4+c^2)}+\dfrac{1-d}{(4+d)(4+d^2)}+\dfrac{1-e}{(4+e)(4+e^2)} \geq 0.$$

Suppose that $a \geq b \geq c \geq d \geq e$. Then, we get $$\dfrac{1-a}{4+a} \leq \dfrac{1-b}{4+b} \leq \dfrac{1-c}{4+c} \leq \dfrac{1-d}{4+d} \leq \dfrac{1-e}{4+e}.$$ and $$\dfrac{1}{4+a^2} \leq \dfrac{1}{4+b^2} \leq \dfrac{1}{4+c^2} \leq \dfrac{1}{4+d^2} \leq \dfrac{1}{4+e^2}.$$

Applying the Chebyshev inequality, one gets $$ \sum_{cyc}\dfrac{1-a}{(4+a)(4+a^2)} \geq \dfrac{1}{5} \sum_{cyc}\dfrac{1-a}{4+a}. \sum_{cyc}\dfrac{1}{4+a^2} = \dfrac{1}{5}\sum_{cyc}\dfrac{1}{4+a^2} \sum_{cyc} \left( \dfrac{5}{4+a}-1 \right)=0.$$

Answer 1

Another way.

By AM-GM $$\sum_{cyc}\frac{a}{4+a^2}\leq\sum_{cyc}\frac{a}{2a+3}.$$ Thus, it's enough to prove that: $$1-\sum_{cyc}\frac{a}{2a+3}\geq0$$ or $$\sum_{cyc}\left(\frac{1}{5}-\frac{a}{2a+3}\right)\geq0$$ or $$\sum_{cyc}\frac{1-a}{2a+3}\geq0$$ or $$\sum_{cyc}\left(\frac{1-a}{2a+3}+5\left(\frac{1}{5}-\frac{1}{4+a}\right)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2}{(2a+3)(4+a)}\geq0.$$

Answer 2

We need to prove that $$1-\sum_{cyc}\frac{a}{4+a^2}\geq0$$ or $$\sum_{cyc}\left(\frac{1}{5}-\frac{a}{4+a^2}+3\left(\frac{1}{5}-\frac{1}{4+a}\right)\right)\geq0$$ or $$\sum_{cyc}\frac{a^3-a^2-a+1}{(4+a^2)(4+a)}\geq0.$$ Now, by AM-GM $$a^3+\frac{1}{2}\geq\frac{3}{2}a^2$$ and $$\frac{1}{2}a^2+\frac{1}{2}\geq a,$$ which after summing gives $$a^3-a^2-a+1\geq0$$ and we are done!