Help in proving $\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{a(bc+1)}$ I was practising inequality problems from an online handout. I stumbled across this inequality problem which seems easy to solve but I wasn't able to move anywhere while trying to solve it;

Prove that for all real numbers $a,b,c\geq 1,$ we have $$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{a(bc+1)}.$$

I tried squaring both sides and then solving, but it only got worse. I was able to progress using calculus, but calculus shouldn't be needed in such contest problems. I checked it on WolframAlpha, and found that it is true. A hint or a solution would be appreciated. Thanks in advance :)
 A: Hello NiceMathophile welcome to M.S.E.
Here's a hint to solve the problem;

Can you show that $\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{bc},$ and then show that $\sqrt{a-1}+\sqrt{bc}\leq \sqrt{a(bc+1)}$?


Use Cauchy Schwartz inequality in both cases.

If you are done solving the problem, you can take a look at the answer below (spoiler alert);

 We know that this inequality is non-symmetric in general which makes it hard to try and solve, but it is symmetric in $b$ and $c$. So we will first find a bound on $\sqrt{b-1}+\sqrt{c-1},$ and then approach to solve the entire problem. By the Cauchy Schwartz inequality, we have;\begin{align*}\left(\sqrt{(b-1)\cdot 1}+\sqrt{1\cdot(c-1)}\right)^2 &\leq \left((b-1)+1\right)\left(1+(c-1)\right)\\ & \leq bc.\end{align*} Now, taking the Square Root gives us;\begin{align*}\sqrt{b-1}+\sqrt{c-1} &\leq bc.\end{align*} Now, it suffices to show that;\begin{align*}\sqrt{a-1}+\sqrt{bc}\leq \sqrt{a(bc+1)}\end{align*} But by the Cauchy Schwartz inequality, we have;\begin{align*}&\left(\sqrt{(a-1)\cdot 1}+\sqrt{1\cdot bc}\right)^2\leq {\left((a-1)+1\right)(bc+1)}\\ \implies &\sqrt{a-1}+\sqrt{bc}\leq {a(bc+1)}.\end{align*}Thus, bringing it all together, we get that;$$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq {a(bc+1)},$$ as required.

A: Try to apply the Cauchy-Schwarz inequality in the following way:
$$
\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}=\sqrt{a-1}\cdot 1+1\cdot (\sqrt{b-1}+\sqrt{c-1})\le
\\
\le\sqrt{(\sqrt{a-1})^2+1^2}\cdot\sqrt{1^2+(\sqrt{b-1}+\sqrt{c-1})^2}.
$$
Can you continue now?
