If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$ 
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$

I just learnt to prove If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$ a few days ago, and it was  just basic application of CS. But I find this one really difficult. I can not apply CS here directly on the numbers $a+\frac{1}{a}$ and $b+\frac{1}{b}$ because CS is for squares. So some manipulation is needed. Anything from hint to full answer will be appreciated.

An alternative approach: $h(x)=x+\frac{1}{x}$ is positive and log-convex over $(0,1)$, since 
$$\frac{d^2}{dx^2}\log h(x)=\frac{(1+3x^2)+(x^2-x^4)}{(x+x^3)^2}>0.$$
In particular $h(x)^3$ is convex and the claim follows from Jensen inequality, as already remarked by Micheal Rozenberg.
