Let $t, a, b>0$ and $n\in\mathbb{N}$. Is it true that
$$|\left(t^2+a^2\right)^{\frac{n+1}{2}} - \left(t^2+b^2\right)^{\frac{n+1}{2}}|\leq t^{\frac{n-1}{2}} |a+b|^{\frac{n+1}{2}}|a-b|^{\frac{n+1}{2}}?$$
Note that for $n=1$, this is true, since $|a^2-b^2| = |a-b||a+b|$. (This desired inequality appeared in one of my calculations in harmonic analysis.)
Does anyone know if this is true? If it is false, then what $t, a, b>0$ form a counter example?
When $t \leqslant 1$, and $n=2k-1,k\geqslant2$, $a \neq b$, we can assume that $$|(t^2+a^2)^k-(t^2+b^2)^k|\leqslant t^{k-1}|a+b|^k|a-b|^k$$ $$\Rightarrow |a^{2k}-b^{2k}+\sum_{i=1}^{k}{{k \choose i}t^{2i}(a^{2(k-i)}-b^{2(k-i)})}|\leqslant t^{k-1}|(a^2-b^2)^k|$$ Notice that, if $a>b$, and $n$ is positive integer$$(a-b)^n \leqslant a^n-b^n$$ $$\therefore |a^{2k}-b^{2k}+\sum_{i=1}^{k}{{k \choose i}t^{2i}(a^{2(k-i)}-b^{2(k-i)})}|\leqslant t^{k-1}|a^{2k}-b^{2k}|$$ $$\Rightarrow |1+{{\sum_{i=1}^{k}{{k \choose i}t^{2i}(a^{2(k-i)}-b^{2(k-i)})}}\over{a^{2k}-b^{2k}}}|\leqslant t^{k-1}$$ But according to assumption we can easily know that $LHS>1$ and $RHS\leqslant 1$. Therefore, the inequality is not true.
Example:$t=1,a=3,b=2,n=3$