Prove by Lemma $1 . x>0 \quad x^{a}-\alpha x \leqslant 1-\alpha(01$
and $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=\frac{1}{\mathrm{p}}+\frac{\mathrm{p}-1}{\mathrm{p}}=1$
Secondly, for $x^{a} \leqslant \alpha x+1-\alpha \quad(x>0,00, b>0, \frac{a^{p}}{b^{q}}>0)$ we get $\left(\frac{a^{p}}{b^{q}}\right)^{\frac{1}{p}} \leqslant \frac{1}{p} \frac{a^{p}}{b^{q}}+\frac{1}{q}$
that is
$$\frac{a}{(b)_{p}^{q}} \leqslant \frac{1}{p} \frac{a^{p}}{b^{q}}+\frac{1}{q}$$

Now calculate $b \cdot b_{p}^{\frac{q}{p}}=$ ?
By $b \cdot b_{p}^{q}=b^{1+\frac{q}{p}}=b^{\frac{p+g}{p}}$
and by $\quad \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=1 \quad \therefore \mathrm{p}+\mathrm{q}=\mathrm{pq}$
Therefore $b \cdot b_{p}^{q}=b^{\frac{p+q}{q}}=b_{p}^{\frac{p q}{p}}=b^{q}$
Hence $b \cdot b_{p}^{g}=b^{q}$
(2) Multiply both sides
$$b \cdot b_{p}^{g}=b^{q}$$

We get $b \cdot b_{p}^{\frac{q}{p}}=b^{q} \quad \frac{a}{b_{p}^{q}} \leqslant\left[\frac{1}{p} \frac{a^{p}}{b^{q}}+\frac{1}{q}\right]$
That is $a b \leqslant \frac{1}{p} a^{p}+\frac{1}{q} b^{q}$