how to obtain this inequality $|\langle A_{t}(x),x\rangle| \leq C( f_{t}+f_{t}^\frac{p_{0}}{2}+(1+|x|_{H})^{p_{0}}+|x|^{\alpha}_{V}+|x|^{\alpha}_{V}+|x|^{\alpha}_{V}|x|^{\beta}_{H})$. I am reading a paper that has following assumption. Using these assumption it has shown above inequality but I am not able to do that please help. We assume that $\alpha>0,p_{0}\geq\beta+2,\theta>0,K,L$ and $f^{\frac{p_{0}}{2}}\in L^{\frac{p_{0}}{2}}((0,T)X\Omega; \mathbb{R})$,such that the following conditions hold almost surely for all $t \in [0,T]:$
We assume that H is a separable Hilbert space, V is a separable, reflexive Banach space embedded continuously and densely in H.
A2)Local Monotonicity :For all $ x,\bar{x}$ in V
$2\langle A_{t}(x)-A_{t}(\bar{x}),x-\bar{x}\rangle +\sum_{j=1}^{\infty}|B^{j}_{t}(x)-B^{j}_{t}(\bar{x})|^{2}_{H}\leq L (1+|\bar{x}|^{\alpha}_{V})(1+|\bar{x}|^{\beta}_{H})|x-\bar{x}|^{2}_{H}$
$\implies$ $2\langle A_{t}(x)-A_{t}(\bar{x}),x-\bar{x}\rangle +\sum_{j=1}^{\infty}|B^{j}_{t}(x)-B^{j}_{t}(\bar{x})|^{2}_{H}\leq L (1+|\bar{x}|^{\alpha}_{V}+|\bar{x}|^{\beta}_{H}+|\bar{x}|^{\alpha}_{V}|\bar{x}|^{\beta}_{H})|x-\bar{x}|^{2}_{H}$
A3)Coercivity: For $x\in V$ , $2\langle A_{t}(x),x\rangle +(p_{0}-1)\sum_{j=1}^{\infty}|B_{t}^{j}(x)|^{2}_{H}+\theta|x|^{\alpha}_{V}\leq f_{t}+K|x|^{2}_{H}$
A4)Growth of A: For all $ x \in V$,
$|A_{t}(x)|^{\frac{\alpha}{\alpha-1}}_{V^{*}}\leq(f_{t}+K|x|^{\alpha}_{V})(1+|x|^{\beta}_{H})$
$\implies |A_{t}(x)|^{\frac{\alpha}{\alpha-1}}_{V^{*}}\leq(f_{t}+f_{t}|x|^{\beta}_{H}+K|x|^{\alpha}_{V}+K|x|^{\alpha}_{V}|x|^{\beta}_{H})$
Now,
$|\langle A_{t}(x),x\rangle|\leq |A_{t}(x)|$ $|x|$ (By Holder's Inequality)
$\implies|\langle A_{t}(x),x\rangle| \leq \frac{\alpha-1}{\alpha}|A_{t}(x)|^{\frac{\alpha}{\alpha-1}}_{V^{*}}+\frac{1}{\alpha}|x|^{\alpha}_{V}$ (By Young's Inequality) $\implies|\langle A_{t}(x),x\rangle|\leq \frac{\alpha-1}{\alpha}(f_{t}+K|x|^{\alpha}_{V})(1+|x|^{\beta}_{H})+\frac{1}{\alpha}|x|^{\alpha}_{V}$(By A4) $\implies |\langle A_{t}(x),x\rangle| \leq C( f_{t}+f_{t}^\frac{p_{0}}{2}+(1+|x|_{H})^{p_{0}}+|x|^{\alpha}_{V}+|x|^{\alpha}_{V}+|x|^{\alpha}_{V}|x|^{\beta}_{H})$
This is my work /my way of solving second last inequality to get the above form of last inequality but not able to do so . Consider,
i)1 < $\alpha \implies\alpha-1>0$
ii)1<$\alpha$
$\implies \frac{1}{\alpha}<1$
By multiplying $\alpha-1$ on both sides of inequality as $\alpha-1>0$ (so sign of inequality will not not change ),
$\implies \frac{\alpha-1}{\alpha}<\alpha-1$
$\implies|\langle A_{t}(x),x\rangle| \leq \frac{\alpha-1}{\alpha}(f_{t}+f_{t}|x|^{\beta}_{H}+K|x|^{\alpha}_{V}+K|x|^{\alpha}_{V}|x|^{\beta}_{H})+\frac{1}{\alpha}|x|^{\alpha}_{V}$(By A4)
$\implies|\langle A_{t}(x),x \rangle | \leq (\alpha-1)(f_{t}+f_{t}|x|^{\beta}_{H}+K|x|^{\alpha}_{V}+K|x|^{\alpha}_{V}|x|^{\beta}_{H})+|x|^{\alpha}_{V}$(By i,ii)
$\implies|\langle A_{t}(x),x\rangle| \leq (\alpha-1)f_{t}+(\alpha-1)f_{t}|x|^{\beta}_{H}+(K(\alpha-1)-1)|x|^{\alpha}_{V} +K(\alpha-1)|x|^{\alpha}_{V}|x|^{\beta}_{H}$
Take $C=sup\{(\alpha-1),(K(\alpha-1)-1),K(\alpha-1)\}$
$\implies|\langle A_{t}(x),x\rangle|\leq C( f_{t}+f_{t}|x|^{\beta}_{H}+|x|^{\alpha}_{V}+|x|^{\alpha}_{V}|x|^{\beta}_{H})$