Hello, it is a right angle triangle. You may consider the substitution $a=x+y, b=y+z, c=z+x$

Let $r:=c$ and apply the substitutions $a=rs$ and $b=rc$, where $s:=\sin x$ and $c:=\cos x$ (and $x$ defined such that $s,c>0$). Then \[\frac{a^3+b^3+r^3}{abr}=\frac{s^3+c^3+1}{sc}.\] Differentiating by the quotient rule and setting equal to $0$ yields \[\frac{(3s^2c^2+(s+c)(s^3+c^3+1))(s-c)}{s^2c^2}=0\] \[\Leftrightarrow (3s^2 c^2+(s+c)(s^3+c^3+1))(s-c)=0.\] But since $s,c>0$ the first factor is positive, so we must have $s=c$. Now $s,c>0$ implies that $x$ (from $\sin x,\cos x$) lies in the first quadrant, so $s=c$ implies that $x=45^o$. Hence min occurs when $a=b=r\sin 45^o=r/\sqrt{2}$, giving \[\frac{a^3+b^3+r^3}{abr}=\frac{s^3+c^3+1}{sc}\ge \frac{2(1/\sqrt{2})^3+1}{(1/\sqrt{2})^2}=\boxed{2+\sqrt{2}}.\]

Thanks. Let $a, b, c$ be side lengths of a right triangle and $c$ be the length of the hypotenuse .Find the minimum value of $\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}$.

Maybe this helps $ \frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc} = \sin x + \cos x + \tan x +%Error. "cosec" is a bad command. x + \sec x + \cot x$

See PE23 from here.

sqing wrote: Thanks. Let $a, b, c$ be side lengths of a right triangle and $c$ be the length of the hypotenuse .Find the minimum value of $\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}$. \[\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}=\frac{a+b}{\sqrt{a^2+b^2}}+\frac{a^2+b^2}{ab}+\frac{(a+b)\sqrt{a^2+b^2}}{ab}=\] \[\left[\frac{a+b}{\sqrt{a^2+b^2}}+\frac{(a+b)\sqrt{a^2+b^2}}{2ab}\right]+\frac{a^2+b^2}{ab}+\frac{(a+b)\sqrt{a^2+b^2}}{2ab}\ge\] \[\frac{\sqrt 2(a+b)}{\sqrt{ab}}+2+\frac{2\sqrt{ab}\sqrt{2ab}}{2ab}\ge 2+3\sqrt 2. \]

sqing wrote: Let $a, b, c$ be side lengths of a right triangle and $c$ be the length of the hypotenuse .Find the minimum value of $\frac{a^3+b^3+c^3}{abc}$. $\frac{a^3+b^3+c^3}{abc}=\frac{a^3+b^3+(a^2+b^2)c}{abc}\geq \frac{\frac{1}{2}(a^2+b^2)(a+b)+(a^2+b^2)c}{abc}$ $=\frac{\frac{1}{2}\sqrt{a^2+b^2}(a+b)+a^2+b^2}{ab}\geq \frac{\sqrt{ 2}ab+2ab}{ab}=2+\sqrt 2. $ here here here

Let $a,b,c$ be positive numbers such that $a^2=b^2+c^2$. Prove that$$\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 5+3\sqrt{2}$$

Since this is a right triangle, we may, WLOG, let $a=1, b=\sin{x}, c=\cos{x}.$ Expaning and cancelling terms we are left to show that $$\sin{x}+\cos{x}+\frac{1}{\sin{x}}+\frac{1}{\cos{x}}+\frac{\cos{x}}{\sin{x}}+\frac{\sin{x}}{\cos{x}} \ge 2+3\sqrt{2} \ \ \ \forall x \in I:=\bigg(0, \frac{\pi}{2}\bigg)$$ Define $f(x)=\sin{x}+\cos{x}+\frac{1}{\sin{x}}+\frac{1}{\cos{x}}.$ Notice that $f'(x)=\frac{\sin^5{x}-\cos^5{x}}{\sin^2{x}\cos^2{x}}.$ $f$ approaches $+\infty$ at either endpoints of $I$. Furthermore, $f$ is decreasing on $\bigg(0, \frac{\pi}{4}\bigg)$ and increasing on $\bigg(\frac{\pi}{4}, \frac{\pi}{2}\bigg).$ Hence, $f(x)$ attains its minimum at $x=\frac{\pi}{4}. $ $f(x) \ge 3\sqrt{2}.$ By AM-GM, $\frac{\cos{x}}{\sin{x}}+\frac{\sin{x}}{\cos{x}} \ge 2$. Combining these two yields the result

Let $a,b,c$ be positive numbers such that $a^2=b^2+c^2$. Find the minimum value of $\left(a+b+c\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right).$