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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.

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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.

$$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3=5+\left(\frac{1}{ab}-1\right)^3-3ab\geq5+(4-1)^3-\frac{3}{4}=\frac{125}{4}.$$

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    $\begingroup$ The first term is squared, not cubed. $\endgroup$ – marty cohen Oct 28 '17 at 16:16
  • $\begingroup$ Let me learn this Holder first from wikipedia! $\endgroup$ – ami_ba Oct 28 '17 at 16:17
  • $\begingroup$ @marty cohen It's a typo in the given. $\endgroup$ – Michael Rozenberg Oct 28 '17 at 16:18
  • $\begingroup$ i think Michael, you have misreaded the given problem $\endgroup$ – Dr. Sonnhard Graubner Oct 28 '17 at 16:18
  • $\begingroup$ ok it was corrected right now $\endgroup$ – Dr. Sonnhard Graubner Oct 28 '17 at 16:19
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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.

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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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