Minimum length of the hypotenuse 
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Prove that the minimum length of the hypotenuse is $(a^{2/3}+b^{2/3})^{3/2}$.


My Attempt
$\frac{x}{y}=\frac{a}{CM}=\frac{AN}{b}$
$$
\frac{x}{y}=\frac{AN}{b}\implies y=\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h(x)=x+y=x+\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h'(x)=1+\frac{\sqrt{x^2-a^2}.b-xb.\frac{x}{\sqrt{x^2-a^2}}}{x^2-a^2}=1+\frac{x^2b-a^2b-x^2b}{(x^2-a^2)^{3/2}}\\
=1+\frac{-a^2b}{(x^2-a^2)^{3/2}}=\frac{(x^2-a^2)^{3/2}-a^2b}{(x^2-a^2)^{3/2}}
$$
$$
h'(x)=0\implies (x^2-a^2)^{3/2}=a^2b\implies (x^2-a^2)^{3}=a^4b^2\\
\implies x^6-3x^4a^2+3x^2a^4-a^6=a^4b^2\implies x^6-3x^4a^2+3x^2a^4-a^6-a^4b^2=0\\
$$
How do I proceed further and find $h_{min}$ without using trigonometry ? Or is there anything wrong with my calculation ?
 A: Following your solution from this point:
$$(x^2-a^2)^{3/2}=a^2b \Rightarrow x^2=(a^2b)^{2/3}+a^2 \Rightarrow x=a^{2/3}(a^{2/3}+b^{2/3})^{1/2}.$$
So:
$$y=\frac{xb}{\sqrt{x^2-a^2}}=\frac{a^{2/3}(a^{2/3}+b^{2/3})^{1/2}b}{\sqrt{((a^2b)^{2/3}+a^2)-a^2}}=(a^{2/3}+b^{2/3})^{1/2}b^{2/3}.$$
Hence:
$$x+y=a^{2/3}(a^{2/3}+b^{2/3})^{1/2}+(a^{2/3}+b^{2/3})^{1/2}b^{2/3}=\\
(a^{2/3}+b^{2/3})^{1/2}(a^{2/3}+b^{2/3})=(a^{2/3}+b^{2/3})^{3/2}$$
A: we have $$\cos(\alpha)=\frac{a}{x}$$ and $$\sin(\alpha)=\frac{b}{y}$$ so we get
$$1=\frac{a^2}{x^2}+\frac{b^2}{y^2}$$ with this equation you ca eliminate $x$ or $y$ in $$c=x+y$$
A: Let $\angle PCB=\angle ABN=\theta$. Then $\displaystyle \cos \theta =\frac{a}{x}$ and $\displaystyle \sin \theta=\frac{b}{y}$.
Then we have to minimize $$x+y=\frac{a}{\cos \theta}+\frac{b}{\sin \theta}$$
So using Holder,s Inequality
$$(\cos^2\theta+\sin^2\theta)\cdot \bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bigg)\cdot \bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bigg)\geq \bigg(a^{\frac{2}{3}}+b^{\frac{2}{3}}\bigg)^3$$
So $$\bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bigg)\geq \bigg(a^{\frac{2}{3}}+b^{\frac{2}{3}}\bigg)^{\frac{3}{2}}$$
