<s>[INST] How can we prove this inequality? Let $a$ , $b$ and $c$ be positive real numbers and $a+b+c=1$ How can we show this inequality?
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
Thanks.
 [/INST] We have that
$$1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$$
so it suffices to show that
$$ab+bc+ca\geq \sqrt{3abc}.$$
Now $X^2+Y^2+Z^2\geq XY+YZ+XZ$ implies
$$ab+bc+ca=(\sqrt{ab})^2+(\sqrt{bc})^2+(\sqrt{ca})^2
\\\geq b\sqrt{ac}+c\sqrt{ab}+a\sqrt{bc}
=(\sqrt{a}+\sqrt{b}+\sqrt{c})\sqrt{abc}.$$
Hence, we still have to prove that
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{3}$$
which holds because $\sqrt{x}$ is concave and
$$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\geq \sqrt{\frac{a+b+c}{3}}=\frac{1}{\sqrt{3}}.$$
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