soryn wrote: If $a,b,c>0,a+b+c=1$,then: $\frac{1}{abc}+\frac{4}{a^{2}+b^{2}+c^{2}}\geq\frac{13}{ab+bc+ca}$ But it's a trivial checking by $uvw$

The next is stronger and another case of equality If $a,b,c>0$, $a+b+c=1$,then: $\frac{1}{abc}+\frac{4\sqrt2}{a^{2}+b^{2}+c^{2}}\geq\frac{9+4\sqrt2}{ab+bc+ca}$

soryn wrote: If $a,b,c>0,a+b+c=1$,then: $\frac{1}{abc}+\frac{4}{a^{2}+b^{2}+c^{2}}\geq\frac{13}{ab+bc+ca}$ Denote $$p=a+b+c, q=ab+bc+ca, r=abc$$The inequality is equivalent with $$q+30rq\geq13r+2q^2$$, which is true, using that $$r\leq \frac{1}{27}, q\leq \frac{1}{3}, q\geq 3\sqrt[3]{r^2}\geq 9r.$$

@above Maybe I'm missing something, but your last step isn't very clear to me. I'm not sure how you can compare the $30qr$ with either $c\cdot r$ or $c\cdot q^2$ (for some constant $c$) using the inequalities you mentioned. Since I flunked the NMO for my grade level anyway, here I go (I'd appreciate someone checking my work, it's highly possible that I miscomputed something and got it all wrong). Using the same notations as above, we homogenize, rewriting the given as $$p^3q+30qr\geq 13p^2r+2pq^2$$$$(p^2-2q)(pq-9r)\geq 4r(p^2-3q)$$$$\sum (a^2+b^2+c^2)\cdot a\cdot (b-c)^2\geq \sum 2abc\cdot (b-c)^2$$$$\sum (b-c)^2\cdot a\cdot (a^2+b^2+c^2-2bc)\geq 0$$$$\sum (b-c)^2\cdot a\cdot \left[a^2+(b-c)^2\right]\geq 0$$which is clearly true. Equality is reached if and only if $a=b=c=\frac{1}{3}$.

We can rewrite the inequality as $(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)+\frac{4\sum ab}{\sum a^2}\geq 13.$ Since $(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)=9+\sum\frac{(a-b)^2}{ab}\geq 9+\frac{2\sum(a-b)^2}{\sum a^2}=9+\frac{4(\sum a^2-\sum ab)}{\sum a^2}= 13-\frac{4\sum ab}{\sum a^2}$ we obtain the desired inequality. Equality holds iff $a=b=c=\frac{1}{3}$

Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$\frac{2}{abc}+\frac{1}{a^3+b^3+c^3}\geq\frac{21}{ab+bc+ca}$$$$\frac{1}{abc}+\frac{3}{a^3+b^3+c^3}\geq\frac{18}{a^2+b^2+c^2}$$

Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$\frac{3}{abc}+\frac{1}{a^4+b^4+c^4}\geq\frac{36}{a^2+b^2+c^2}$$

MariusStanean wrote: The next is stronger and another case of equality If $a,b,c>0$, $a+b+c=1$,then: $$\frac{1}{abc}+\frac{4\sqrt2}{a^{2}+b^{2}+c^{2}}\geq\frac{9+4\sqrt2}{ab+bc+ca}$$ Mathematical Reflections 1 (2023) S618

MariusStanean wrote: The next is stronger and another case of equality If $a,b,c>0$, $a+b+c=1$,then: $\frac{1}{abc}+\frac{4\sqrt2}{a^{2}+b^{2}+c^{2}}\geq\frac{9+4\sqrt2}{ab+bc+ca}$ pqr method: Let $p = a + b + c = 1, q = ab + bc + ca, r = abc$. It suffices to prove that $$\frac{1}{r} + \frac{4\sqrt 2}{1 - 2q} \ge \frac{9 + 4\sqrt 2}{q}.$$ Using $p^2 \ge 3q$, we have $0 < q \le 1/3$. Let $t = \sqrt{1 - 3q} \in [0, 1)$. We have $q = \frac{1}{3}(1 - t^2)$. Using $$0 \le (a - b)^2(b - c)^2(c - a)^2 = - 4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2,$$we have $$-729r^2 + (-162t^2 + 54)r + 4t^6 - 9t^4 + 6t^2 - 1 \ge 0$$which results in $$r \le \frac{2}{27}t^3 - \frac19 t^2 + \frac{1}{27}.$$ It suffices to prove that $$\frac{1}{\frac{2}{27}t^3 - \frac19 t^2 + \frac{1}{27}} + \frac{4\sqrt 2}{1 - 2\cdot \frac{1}{3}(1 - t^2)} \ge \frac{9 + 4\sqrt 2}{\frac{1}{3}(1 - t^2)}$$which is true.