Four years ago, I proposed an inequality related to area and sides of a polygon

Now, In my computer checked, the Inequality above can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4\tan\frac{\pi}{n}Area(A_1A_2\cdots A_n) \le \sum_{i=1}^nA_iA_{i+1}^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Question: May you give your proof of the conjecture inequality above?

PS: Using my computer I also found the formula of the area of equalangular convex pentagon as follows:

Let $ABCDE$ be a equalangular convex pentagon then $Area(ABCDE)=\frac{1}{4}\cot\frac{\pi}{5}(a^2+b^2+c^2+d^2+e^2-\frac{(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2}{\sqrt{5}})$

Four years ago, I proposed an inequality related to area and sides of a polygon.

After computer checking, I conjecture that the previous inequality can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4(\tan\frac{\pi}{n})Area(A_1A_2\cdots A_n) \le \sum_{i=1}^n(A_iA_{i+1})^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Question: Can you prove this conjectured inequality?

PS: I also found that for an equiangular convex pentagon, $Area(ABCDE)=\frac{1}{4}(\cot\frac{\pi}{5})(a^2+b^2+c^2+d^2+e^2-\frac{(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2}{\sqrt{5}})$

Four years ago, I proposed an inequality related to area and sides of a polygon

Now, In my computer checked, the Inequality above can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4\tan\frac{\pi}{n}Area(A_1A_2\cdots A_n) \le \sum_{i=1}^nA_iA_{i+1}^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Question: May you give your proof of the conjecture inequality above?

Four years ago, I proposed an inequality related to area and sides of a polygon

Now, In my computer checked, the Inequality above can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4\tan\frac{\pi}{n}Area(A_1A_2\cdots A_n) \le \sum_{i=1}^nA_iA_{i+1}^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Question: May you give your proof of the conjecture inequality above?

PS: Using my computer I also found the formula of the area of equalangular convex pentagon as follows:

Let $ABCDE$ be a equalangular convex pentagon then $Area(ABCDE)=\frac{1}{4}\cot\frac{\pi}{5}(a^2+b^2+c^2+d^2+e^2-\frac{(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2}{\sqrt{5}})$

Four years ago, I proposed an inequality related to area and sides of a polygon

Now, In my computer checked, the Inequality above can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4\tan\frac{\pi}{n}Area(A_1A_2\cdots A_n) \le \sum_{i=1}^nA_iA_{i+1}^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Equality holds iff $A_1A_2\cdots A_n$ is a regular n-gon.

Question: May you give your proof?

Four years ago, I proposed an inequality related to area and sides of a polygon

Now, In my computer checked, the Inequality above can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4\tan\frac{\pi}{n}Area(A_1A_2\cdots A_n) \le \sum_{i=1}^nA_iA_{i+1}^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Question: May you give your proof of the conjecture inequality above?