Doubt on Lebesgue integration How to prove that $\int_{\Omega}(a+b|u|^{(p-1)})^{q}\le C\int_{\Omega}(1+|u|^{(p-1)q})$ where $ u\in L^{p}(\Omega)$, and a,b are constants. q is a conjugate exponent  of p
Note:
This answer may show how little I know
about Lebesgue integration.
I am treating this just as
an inequality between the integrands
and assuming some bounds on the parameters.

This is true if
$(1+x^u)^v
&lt; C(1+x^{uv})
$,
where
$x &gt; 0$
and
$C$ depends on $u$ and $v$
and $u = [-1$ and $v = q$.

If we let $x^u = y$,
this becomes
$(1+y)^v &lt; C(1+y^v)$,
or
$1+y &lt; C(1+y^v)^{1/v}$,
where the $C$'s are different
and depend on $u$ and $v$.

By the arithmetic means inequality,
if $v &gt; 1$,
$\dfrac{1+y}{2}
&lt; \left(\dfrac{1+y^v}{2}\right)^{1/v}
$
or
$1+y
&lt; 2^{1-1/v}(1+y^v)^{1/v}
$

So, if $p&gt;1$ and $q &gt; 1$,
this seems to be true.
Note:
This answer may show how little I know
about Lebesgue integration.
I am treating this just as
an inequality between the integrands
and assuming some bounds on the parameters.

This is true if
$(1+x^u)^v
&lt; C(1+x^{uv})
$,
where
$x &gt; 0$
and
$C$ depends on $u$ and $v$
and $u = [-1$ and $v = q$.

If we let $x^u = y$,
this becomes
$(1+y)^v &lt; C(1+y^v)$,
or
$1+y &lt; C(1+y^v)^{1/v}$,
where the $C$'s are different
and depend on $u$ and $v$.

By the arithmetic means inequality,
if $v &gt; 1$,
$\dfrac{1+y}{2}
&lt; \left(\dfrac{1+y^v}{2}\right)^{1/v}
$
or
$1+y
&lt; 2^{1-1/v}(1+y^v)^{1/v}
$

So, if $p&gt;1$ and $q &gt; 1$,
this seems to be true.