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The problem can be equivalently stated as

PROBLEM : A convex $n$ sided polygon has a circumcircle and an inscribed circle, its area is $B$, and the areas of circumcircle and inscribed circle are $A$ and $C$ respectively. Prove that $2B < A+C$.

I think this problem is very difficult. This is my attempt for special case of polygons i.e Regular polygons.

Naming of parameters :

$R$ be the radius of circumcircle of the polygon.
$r$ be inradius of the polygon.
$n$ be the number of sides of polygon. $\theta$ = $\frac{2\pi}{n}$ = angle subtended by a side of polygon at centre.
$a$ be the length of side of polygon.

Relations between $R,r,a,\theta$ :

$R^2 = \frac{a^2}{4} + r^2$, $a = 2R*sin(\frac{\theta}{2})$ and $r = R*cos(\frac{\theta}{2})$

We need to prove $2B < A+C$
$\Leftrightarrow \frac{2sin(\theta)}{3+cos(\theta)} < \frac{\pi}{n}$

This can be verified by showing that, inequality is true for $n = 3 $ and LHS decreases faster than RHS.

The method I used for regular polygons isn't applicable to all. There is too much freedom and ambiguity. But I don't have any idea to tackle generalized polygon. Can anybody help me?

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It's actually simple. If the perimeter of your convex shape is $P$, then your inequality is equivalent to: $$\pi(r^2+R^2) > Pr,$$ where $R$ and $r$ are the circumradius and inradius, respectively. A trivial upper bound on the perimeter is: $$P < 2\pi R$$ where the latter is the length of the circumcircle. But now a simple AM-GM finishes the problem:

$$Pr < 2\pi Rr < \pi(r^2+R^2)$$

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    $\begingroup$ ........No words to say. Beautiful proof. I unnecessarily overcomplicated the problem. How stupid of me.. Thanks @dezdichado $\endgroup$ Commented Dec 5, 2020 at 18:21
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    $\begingroup$ @MathematicalCuriosity no need to call yourself stupid. This is pure experience - I did probably thousands of Euclidean geometry problems in secondary school for math contest and so I happen to know basic ideas, techniques that are applicable in most cases. $\endgroup$ Commented Dec 5, 2020 at 18:24

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