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.

By Holder and AM-GM we obtain:
$$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3=\frac{1}{4}\cdot(1+1)^2\left(\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3\right)\geq$$
$$\geq\frac{1}{4}\left(a+\frac{1}{a}+b+\frac{1}{b}\right)^3=\frac{1}{4}\left(1+\frac{1}{ab}\right)^3\geq\frac{1}{4}\left(1+\frac{1}{\frac{(a+b)^2}{4}}\right)^3=\frac{125}{4}.$$
Also we can use Jensen here.
Indeed, let $f(x)=\left(x+\frac{1}{x}\right)^3$.
Thus, $$f''(x)=\frac{6(x^6+x^2+2)}{x^5}>0$$ for all $x>0$.
Id est, $$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3\geq2\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^3=\frac{125}{4}.$$
Also we can use AM-GM only.
$$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3=5+\left(\frac{1}{ab}-1\right)^3-3ab\geq5+(4-1)^3-\frac{3}{4}=\frac{125}{4}.$$
