isn't something wrong with the problem? i'll prove that $ \frac{a+c}{\sqrt{a^2+c^2}}< \frac{\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$ it's enough to show that $(a+c)\sqrt{(a^{2}+b^{2})(b^{2}+c^{2})}\leq \sqrt{6} (b+c)^{2}$ use the fact that $\sqrt{(a^{2}+b^{2})}\leq a+b,\sqrt{(c^{2}+b^{2})}\leq c+b$ so we need to prove that $(a+c)(a+b)(b+c)\leq \sqrt{6}(b+c)^{2}$and it's obvious because from $a\leq b\leq c$ we have $a+c\leq b+c,b+c\leq b+c$ and because $a+b+c=1$ so $a+b \leq 1 <\sqrt{6}$. now similarly $ \frac{b+c}{\sqrt{b^2+c^2}}< \frac{\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$ and $ \frac{a+b}{\sqrt{a^2+b^2}}<\frac{\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$ sum them up.so we are done

If we take the natural equality case $ \displaystyle a = b = c = \frac {1} {3} $, then the two sides of the inequality become equal if we get rid of the $ 3\sqrt{6} $ term on the RHS. However, I haven't been able to prove the resulting inequality (and am not even sure if it is true).

math154 wrote: Let $a,b,c$ be three positive real numbers such that $ a \le b \le c$ and $a+b+c=1$. Prove that \[\frac{a+c}{\sqrt{a^2+c^2}}+\frac{b+c}{\sqrt{b^2+c^2}}+\frac{a+b}{\sqrt{a^2+b^2}} \le \frac{3\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}.\] Owen Goff. this can be strengthen to $\frac{a+c}{\sqrt{a^2+c^2}}+\frac{b+c}{\sqrt{b^2+c^2}}+\frac{a+b}{\sqrt{a^2+b^2}} \le \frac{\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}. $ and we know that if x>y then $\frac{{x}\mathrm{{+}}{y}}{\sqrt{{x}^{2}\mathrm{{+}}{y}^{2}}}$ become smaller when x/y get greater so LHS is smaller then $3\frac{{a}\mathrm{{+}}{b}}{\sqrt{{a}^{2}\mathrm{{+}}{b}^{2}}}$ or $3\frac{{c}\mathrm{{+}}{b}}{\sqrt{{c}^{2}\mathrm{{+}}{b}^{2}}}$ the rest is easy and the coefficient $\sqrt{6}$can be reduce again but I don't want to calculate... EDIT in my method the coefficient can be reduce to 1.5 who can do better?