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Here is a classic way of proving this inequality for a triangle. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.

Here is a classic way of proving this inequality for a triangle. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.

Here is a classic way of proving such inequalities and here for a triangle. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.

Here is a classic way of proving this inequality for a triangle. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.

Here is a classic way of proving such inequalities. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.

Here is a classic way of proving such inequalities and here for a triangle. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.