Olympiad Inequality Challenge

[anemone]

Let $a,\,b$ and $c$ be non-negative real numbers such that $a+b+c=1$.

Prove that \(\displaystyle \sum_{cyclic}\sqrt{4a+1} \ge \sqrt{5}+2\).

 

[PaulRS]

Spoiler

To avoid being repetitive, the symbol $\sum$ will denote a cyclic sum.

Observe that
$$
S(a,b,c) = \sum \sqrt{4a+1} = 3 + \sum \left(\sqrt{4a+1} - 1 \right)\,.
$$

The reason we are interested in substracting $1$ is because
$$
\sqrt{4a+1} - 1 = \frac{4a }{\sqrt{4a+1} + 1 }\,,
$$
and here the function $f(x) = \frac{4}{\sqrt{4x+1} + 1 }$ is convex (in preparation for Jensen's inequality), unlike $x\mapsto \sqrt{4x+1}$ (hence the transformation).

Important. Now our target inequality $S(a,b,c)\geq \sqrt{5}+2$ reads $ \sum a f(a) \geq \sqrt{5} -1$.

Since we know $f$ to be convex and $\sum a = 1$, Jensen's inequality now applies and we deduce
$$
\sum a f(a) \geq f(\sum a^2)\,.
$$

Here $\sum a^2$ is not a constant, but $f$ is strictly decreasing. Observe then that $a\geq a^2$ (similarly for $b$ and $c$) because $a,b,c\geq 0$ and $\sum a = 1$ imply $a,b,c\in [0,1]$. Therefore $1\geq \sum a \geq \sum a^2$ $(*)$, hence we get
$$
\sum a f(a) \geq f(\sum a^2) \geq f(1) = \frac{4}{\sqrt{5}+1} = \sqrt{5}-1\,,
$$
as desired.

$(*)$ equality here can hold if and only if $\{a,b,c\} = \{0,1\}$, meaning that one variable equals $1$ and the rest $0$.

 

[anemone]

Awesome, PaulRS! And thanks for participating!

Here is the solution of other that I wanted to share with MHB:

Spoiler

First, note that

$\sqrt{4a+1}+\sqrt{4b+1}\ge 1+\sqrt{4(a+b)+1}$

since

$(\sqrt{4a+1}+\sqrt{4b+1})^2\ge (1+\sqrt{4(a+b)+1})^2$

$4a+1+2\sqrt{4a+1}\sqrt{4b+1}+4b+1\ge 1+2\sqrt{4(a+b)+1}+4(a+b)+1$

$\sqrt{4a+1}\sqrt{4b+1}\ge \sqrt{4(a+b)+1}$

$(4a+1)(4b+1)\ge 4(a+b)+1$

$4ab+4(a+b)+1\ge 4(a+b)+1$ is true for $a,\,b,\,c\in [0,\,1]$.

Therefore we get:

$\sqrt{4a+1}+\sqrt{4b+1}\ge 1+\sqrt{4(a+b)+1}$

$\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\ge 1+\sqrt{4(a+b)+1}+\sqrt{4c+1}$

And

$\begin{align*}\sqrt{4(a+b)+1}+\sqrt{4c+1}&\ge 1+\sqrt{4(a+b)+4c+1}\\&=1+\sqrt{4(a+b+c)+1}\\&=1+\sqrt{4(1)+1}\text{since $a+b+c=1$}\\&=1+\sqrt{5}\end{align*}$

Combining all results the proof is then followed.

$\begin{align*}\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}&\ge 1+\sqrt{4(a+b)+1}+\sqrt{4c+1}\\&\ge 1+1+\sqrt{5}\\&=2+\sqrt{5}\,\,\,\text{Q.E.D.}\end{align*}$

Equality holds when $(a,\,b,\,c)=(0,\,0,\,1)$ and its permutation.