Inequality involving lenghts of a triangle's sides

Question

Let $a,b,c$ be the lengths of a triangle's sides. Show that $$a+ \sqrt{bc} \ge b+c.$$

Answer 1

Counter Example: $a = 1, b = c = 2$. However, the inequality:

$a+2\sqrt{bc} \ge b+c$ is true. To see this, we can assume $b \ge c$. Thus:

$a \ge b+c-2\sqrt{bc} \iff a \ge \left(\sqrt{b}-\sqrt{c}\right)^2\iff \sqrt{a} \ge \sqrt{b} - \sqrt{c}\iff \sqrt{a}+\sqrt{c} \ge \sqrt{b}$, and this last one is true because $\sqrt{b} < \sqrt{a+c} < \sqrt{a}+\sqrt{c}$.

Answer 2

Thanks, @H. H. Rugh and @DeepSea for the replacement of the false inequality proposed by the OP (who gives no news...) by this one,

$$a + 2\sqrt{bc} \ge b+c$$

and its proof.

I propose here an alternative proof. It suffices to set :

$$\begin{cases}a&=&&&q&+&r\\b&=&p&&+&&r\\c&=&p&+&q&&\\\end{cases}$$

for some $p,q,r > 0$. See explanations below.

(almost all issues involving lengthes of sides of triangle can be established in that way ;

Then, with few calculations, the inequality to be proved is shown equivalent to :

$$(p+r)(q+r) \geq r^2$$

which is evidently true, with a "limit case" of equality if $p$ and $q$ tend to $0$, otherwise said when $c$ tend to $0$.

Explanation : the points where the inscribed circle touches the sides divide $a$ into 2 segments with length $q$ and $r$, $b$ into segments with length $p$ and $r$, $c$ into segments with length $p$ and $q$.