parmenides51 wrote: Let $a, b, c$ be side lengths of a right triangle. Determine the minimum possible value of $\frac{a^3 + b^3 + c^3}{abc}$. $ minimum? $ $\frac{a^3 + b^3 + c^3}{abc}\geq \frac{3abc}{abc}=3 $

$$\frac{a^3 + b^3 + c^3}{abc}\geq 2+\sqrt{2}.$$

parmenides51 wrote: Let $a, b, c$ be side lengths of a right triangle. Determine the minimum possible value of $\frac{a^3 + b^3 + c^3}{abc}$. Suppose that $c$ is the hypothenuse, so $c^2=a^2+b^2$. From Holder $2(a^3+b^3)^2 \geq (a^2+b^2)^3$, so $a^3+b^3 \geq \frac{c^3}{\sqrt{2}}$. Also $ab \leq \frac{1}{2}(a^2+b^2)=\frac{1}{2}c^2$ Therefore: $\frac{a^3+b^3+c^3}{abc} \geq \frac{\frac{c^3}{\sqrt{2}}+c^3}{\frac{c^3}{2}}= 2+\sqrt{2}$ Equality when $a =b=\frac{c}{\sqrt{2}}$.

Assume wlog $a^2=b^2+c^2$ Then AM-GM: $a^2=b^2+c^2\ge 2\sqrt{bc}$, power mean inequality $b^3+c^3\ge \frac{1}{\sqrt2}\cdot a^3$ $(b-c)^2+\frac{1}{\sqrt2}\cdot a^2\ge \sqrt2bc$ $a(b-c)^2+\frac{1}{\sqrt2}\cdot a^3\ge \sqrt2abc$ $a(b^2-2bc+c^2)+b^3+c^3\ge \sqrt2abc$ $\frac{a^3 + b^3 + c^3}{abc}\geq 2+\sqrt{2}.$ Equality iff $b=c,a=\sqrt2 b$

Let $a,b,c$ be three positive real numbers such that $a^2=b^2+c^2$ . Prove that$$a^3+b^3+c^3\ge(2+\sqrt{2})abc.$$ $$a^3+b^3+c^3=\left(1-\frac{\sqrt{2}}{4}\right)a^3+\frac{\sqrt{2}}{4}a^3+b^3+c^3\geq2\left(1-\frac{\sqrt{2}}{4}\right)abc+\frac{3\sqrt{2}}{2}abc=(2+\sqrt{2})abc.$$Equality holds when $a=\sqrt{2}b=\sqrt{2}c.$

Let $a, b, c$ be side lengths of a right triangle. Prove that$$\frac{a^4 + b^4 + c^4}{abc(a+b+c)}\geq3(\sqrt 2-1).$$

WLOG let $a> b\geq c$ $a+b\leq \sqrt{2(a^2+b^2)}=\sqrt2 c$ So, $a+b+c\leq (\sqrt{2}+1)c$ $ab\leq \frac{a^2+b^2}{2}=\frac{c^2}{2}$ $a^4+b^4+c^4\geq \frac{(a^2+b^2)^2}{2}+c^4=\frac{3}{2}c^4$ Dividing the third by the first 2, we get the required bound

Let $a, b, c$ be side lengths of a right triangle. Prove that $$\frac{a^5 + b^5 + c^5}{abc(ab+bc+ca)}\geq\frac{8(\sqrt 2-1)}{7}.$$

Let $c$ be the hypotenuse $a+b\leq \sqrt2 c$, as proved earlier $ab\leq\frac{c^2}{2}$ as proved earlier By power mean, $$\sqrt[5]{\frac{a^5+b^5}{2}}\geq \sqrt{\frac{a^2+b^2}{2}}$$, so $$a^5+b^5\geq \frac{c^5}{2\sqrt2}$$So, $LHS=\frac{(a^5+b^5)+c^5}{(ab)(c)(c(a+b)+ab)}$ Substituting the bounds proved above, we get the required bound

Let $a,b,c$ be positive numbers such that $a^2=b^2+c^2$. Prove that$$\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 5+3\sqrt{2}$$\[a^3+b^3+c^3\ge(2+\sqrt{2})abc\]. \[ a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge(2+3\sqrt{2})abc\]. \[a^3+b^3+c^3\ge \frac{2\sqrt{2}+1}{7}[a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)]\]https://artofproblemsolving.com/community/c6h511724p2873311 https://artofproblemsolving.com/community/c6h608170p3615254 https://artofproblemsolving.com/community/c6h608170p3615254 https://artofproblemsolving.com/community/c6h2871334p25507389

Math-wiz wrote: WLOG let $a> b\geq c$ $a+b\leq \sqrt{2(a^2+b^2)}=\sqrt2 c$ So, $a+b+c\leq (\sqrt{2}+1)c$ $ab\leq \frac{a^2+b^2}{2}=\frac{c^2}{2}$ $a^4+b^4+c^4\geq \frac{(a^2+b^2)^2}{2}+c^4=\frac{3}{2}c^4$ Dividing the third by the first 2, we get the required bound Very nice ,but i seems is WLOG let $c> a\geq b$

Suppose that \(c\) is the hypotenuse and \(x\) is an angle between \(c\) and \(b\). Thus, \[ c \sin(x) = a \quad \text{and} \quad c \cos(x) = b. \]Then, substitute to find the minimum of \[ \frac{(\sin(x))^3 + (\cos(x))^3 + 1}{\sin(x)\cos(x)}. \]Consider \[ (\sin(x))^3 + (\cos(x))^3 + 1 = (\sin(x) + \cos(x))(1 - \sin(x)\cos(x)) + 1. \]This yields \[ (\sin(x))^3 + (\cos(x))^3 + 1 \geq \sqrt{2} \left(\frac{1}{2}\right) + 1. \]And \[ \sin(2x) + \frac{\sqrt{2}}{2} \sin(2x) \leq 1 + \frac{1}{\sqrt{2}}. \]Thus, \[ (\sin(x))^3 + (\cos(x))^3 + 1 \geq \sin(2x) + \frac{\sqrt{2}}{2} \sin(2x). \]Rewrite \(\sin(2x)\) as \(2\sin(x)\cos(x)\): \[ \frac{(\sin(x))^3 + (\cos(x))^3 + 1}{\sin(x)\cos(x)} \geq 2 + \sqrt{2}. \]Equality holds when \(a\sqrt{2} = b\sqrt{2} = c\).