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I got this lemma and proof from a book however I cannot understand the proof.enter image description here enter image description here

How can $\alpha\beta\leq\frac{\alpha^p}{p}+\frac{\beta^q}{q}$? What do we take from the last inequality? Sorry for print screening. Thanks for reading

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    The inequality is Young's inequality! – 2017-02-28

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The function $x\mapsto e^x$ is convex so $$ab = \exp(\log a+ \log b)=\exp\left(\frac{1}{p}\log a^p+\frac{1}{q}\log b^q\right)$$ $$\leq\frac{1}{p}e^{\log(a^p)}+\frac{1}{q}e^{\log(b^q)}=\frac{a^p}{p}+\frac{b^q}{q}.$$

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    Could you please provide me an explantation to the last inequality that concludes the proof of the lemma? Anyway thanks for the first proof! – 2017-02-28
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    When you write log you mean $ln$ right? – 2017-02-28
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    @PedroGomes convexity means that $\exp(tx+sy)\leq t\exp(x)+s\exp(y)$ provided $s+t=1$, in our case $s=1/p$ and $t=1/q$. And, yes, $\log =\ln$. Is that okay? – 2017-02-28
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    My doubt also lies in the substitution of $\alpha_i$ and $\beta_i$ by $\frac{\mu_i^p}{p}$${(\Sigma{\mu_j}^p})^\frac{-1}{p}$, $\frac{v_i^p}{p}$${(\Sigma{v_j}^q})^\frac{-1}{q}$ respectively. Could you explain the last inequality? – 2017-02-28
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    Do not mind I already got it. – 2017-03-01