Symmetric matrix power inequality for row sums

The conjecture fails even for a strictly positive definite $2\times2$ matrix.

Let

$$M= \begin{pmatrix} \frac12&\frac12\\[2pt] \frac12&1 \end{pmatrix}, \qquad \alpha=\frac32, \qquad \mathbf1=\binom11.$$

Every entry belongs to $[0,1]$, and

$$\det M=\frac14>0,\qquad \operatorname{tr}M=\frac32,$$

so $M$ is positive definite and its real fractional power is unambiguously defined. The positive square root is

$$M^{1/2} = \frac{M+\frac12I}{\sqrt{5/2}},$$

as follows immediately from the Cayley–Hamilton identity

$$M^2-\frac32M+\frac14I=0.$$

Since

$$M\mathbf1=\binom1{3/2},\qquad \mathbf1^{\mathsf T}M\mathbf1=\frac52,\qquad \mathbf1^{\mathsf T}M^2\mathbf1=\frac{13}{4},$$

we obtain

$$\mathbf1^{\mathsf T}M^{3/2}\mathbf1 = \frac{\frac{13}{4}+\frac12\cdot\frac52}{\sqrt{5/2}} = \frac9{\sqrt{10}}.$$

On the other hand,

$$\mathbf1^{\mathsf T}(M\mathbf1)^{3/2} = 1+\left(\frac32\right)^{3/2} = 1+\frac{3\sqrt6}{4}.$$

These quantities satisfy the strict reverse inequality

$$\frac9{\sqrt{10}} > 1+\frac{3\sqrt6}{4}.$$

Indeed, after multiplying by positive quantities and squaring, this inequality reduces to

$$149>60\sqrt6,$$

which holds because

$$149^2-(60\sqrt6)^2=22201-21600=601>0.$$

Therefore

$$\mathbf1^{\mathsf T}M^\alpha\mathbf1 > \mathbf1^{\mathsf T}(M\mathbf1)^\alpha$$

for the admissible positive-definite matrix $M$ and exponent $\alpha=3/2$.