<s>[INST] 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.
 [/INST] Let $a=\cos^2 t,b=\sin^2 t$. 
Note
$$ \cos^6t+\sin^6t=(\cos^2t+\sin^2t)(\cos^4t-\cos^2t\sin^2t+\sin^4t)=1-3\cos^2t\sin^2t. $$
Then
\begin{eqnarray}
&&(a+\frac1a)^3+(b+\frac1b)^3\\
&=&(\cos^2t+\sec^2t)^3+(\sin^2t+\csc^2t)^3\\
&=&\cos^6t+3\cos^4t\sec^2t+3\cos^2t\sec^4t\\
&&+\sin^6t+3\sin^4t\csc^2t+3\sin^2t\csc^4t+\csc^6t\\
&=&(\cos^6t+\sin^6t)+3(\cos^2t+\sin^2t)+3(\sec^2t+\csc^2t)+\frac{\cos^6t+\sin^6t}{\cos^6t\sin^6t}\\
&=&4-3\cos^2t\sin^2t+\frac{3}{\sin^2t\cos^2t}+\frac{1-3\cos^2t\sin^2t}{\cos^6t\sin^6t}
\end{eqnarray}
Let $x=\cos^2t\sin^2t$ and then $x\in(0,\frac14]$. Define
$$ f(x)=4-3x+\frac3x+\frac{1-3x}{x^3}. $$
Then
$$ f'(x)=-3-\frac3{x^2}+\frac{8x-3}{x^4}<0, x\in(0,\frac12]$$
and hence
$$ f(x)\ge f(\frac12)=\frac{125}{4}. $$
So
$$ (a+\frac1a)^3+(b+\frac1b)^3\ge\frac{125}{4}$$
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