Conjecture: For $a,b,c,A,s$ the sides, area, and semi-perimeter of a triangle, and $x\ge2$: $a+b-c\ge s(2/3)^{(x+1)/(x-1)}(\frac{3^{3/2}A}{s^2})^x$

Question

In the comments of this related question, it was mentioned that the RHS and LHS were expressions in $a,b$ and $c$ so the inequality was not useful. So I wanted to find an improved triangle inequality where for a given RHS, the LHS there are infinitely many possible combinations of $a,b$ and $c$. I obtained following conjecture obtained by empirical means and verified for discrete $2 \le x \le 70$. Can this be proved?

Conjecture: Let $a,b,c$ be the sides of a triangle of area $A$ and semi-perimeter $s$. Then, for any $x \ge 2$,

$$ a+b-c \ge s\left(\frac{2}{3}\right)^{\frac{x+1}{x-1}} \left(\frac{3^{\frac{3}{2}}A}{s^2}\right)^x \tag 1 $$

For a given area $A$ and semi-parameter $s$ there are infinitely many possible values of $a,b$ and $c$ hence the above inequality is gives an explicit lower bound of the triangle inequality.

Question 1: Is the conjecture true?

Question 2: If true, can the inequality be improved?

Note: The difference between LHS and RHS for random $2 \le x \le 70$ was typically or the order of $10^{-8}$ or smaller if $a,b$ and $c$ were the sides of a triangle inscribed in a unit circle.

Update 13-5-2023: For $x \ge 5.23$ experimental data shows that the triangle inequality can be improved to

$$ a+b-c \ge s\left(\frac{2}{3}\right)^{1 + \frac{2}{x}} \left(\frac{3^{\frac{3}{2}}A}{s^2}\right)^x \tag 1 $$

Answer 1

Some thoughts for the conjecture:

Using Ravi's substitution $a = u + v, b = v + w, c = w + u$ for $u, v, w > 0$, we have $$a + b - c = 2v, \tag{1}$$ and $$s = \frac{a + b + c}{2} = u + v + w \tag{2}$$ and $$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(u + v + w)uvw}. \tag{3}$$

The desired inequality becomes $$2v \ge (u + v + w)\left(\frac23\right)^{(x+1)/(x-1)}\left(\frac{27uvw}{(u + v + w)^3}\right)^{x/2}. \tag{4}$$

Using $uw \le \frac{(u+w)^2}{4}$, it suffices to prove that $$2v \ge (u + v + w)\left(\frac23\right)^{(x+1)/(x-1)}\left(\frac{27v(u + w)^2/4}{(u + v + w)^3}\right)^{x/2}. \tag{5}$$

Letting $y = \frac{2v}{u+w}$, it suffices to prove that $$\frac{2y}{y+2} \ge \left(\frac23\right)^{(x+1)/(x-1)}\left(\frac{27y}{(y + 2)^3}\right)^{x/2}. \tag{6}$$ or (taking logarithm) $$f(y) := \ln \frac{2y}{y+2} - \frac{x+1}{x-1}\ln \frac23 - \frac{x}{2}\ln \frac{27y}{(y + 2)^3} \ge 0. \tag{7}$$

If $x = 2$, (7) becomes $\ln\frac{(y+2)^2}{4} \ge 0$ which is true.

If $x > 2$, we have $$f'(y) = \frac{xy - x + 2}{y(y+2)}.$$ Let $y_0 = (x - 2)/x > 0$. We have $f'(y_0) = 0$, and $f'(y) \le 0$ on $(0, y_0)$, and $f'(y) \ge 0$ on $(y_0, \infty)$.

Thus, we have $$f(y) \ge f(y_0) = \ln\frac{2(x-2)}{3x-2} - \frac{x+1}{x-1}\ln \frac23 - \frac{x}{2}\ln \frac{27(x-2)x^2}{(3x-2)^3}.$$ It suffices to prove that, for all $x > 2$, $$\ln\frac{2(x-2)}{3x-2} - \frac{x+1}{x-1}\ln \frac23 - \frac{x}{2}\ln \frac{27(x-2)x^2}{(3x-2)^3} \ge 0.$$

To be continued.

Answer 2

Remarks: We can find the minimum of $a + b - c$ given $s, A > 0$ with $s^4 \ge 27A^2$. This minimum is given by $2u_1$ where $u_1$ is the smallest positive real root of $su(s-u)^2 - 4A^2 = 0$.

Using Ravi's substitution $a = x + y, b = y + z, c = z + x$ for $x, y, z > 0$, we have $$a + b - c = 2y, \tag{1}$$ and $$s = \frac{a + b + c}{2} = x + y + z \tag{2}$$ and $$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x + y + z)xyz}. \tag{3}$$

Fact 1: There exist $x, y, z > 0$ such that $x + y + z = s$ and $\sqrt{(x+y+z)xyz} = A$ if and only if $s, A > 0$ and $s^4 \ge 27A^2$. (The proof of Fact 1 is given at the end.)

By Fact 1, our problem is to find the minimum of $2y$ subject to $x, y, z > 0$; $x + y + z = s$ and $(x+y+z)xyz = A^2$, given $s, A > 0$ with $s^4 \ge 27A^2$.

To this end, we need the following auxiliary result.

Fact 2: Given $s, A > 0$ with $s^4 \ge 27A^2$, the cubic equation $su(s - u)^2 - 4A^2 = 0$ has three positive real roots $0 < u_1\le s/3 \le u_2 < s < u_3$. (The proof of Fact 2 is given at the end.)

By Fact 2, let $u_1$ be the smallest positive real root of $su(s - u)^2 - 4A^2 = 0$.

We claim that the minimum of $2y$ is given by $2u_1$ when $x = z = (s - u_1)/2$ and $y = u_1$.

Proof of the claim: First, using $zx \le (z + x)^2/4$, we have $$A^2 = (x+y+z)xyz \le sy \cdot \frac{(s - y)^2}{4}. \tag{4}$$

Second, using $su_1(s-u_1)^2 - 4A^2 = 0$, we have $$0 \le sy(s-y)^2 - 4A^2 = sy(s-y)^2 - su_1(s-u_1)^2$$ $$= s(y - u_1)[(s - y - u_1)^2 - yu_1]. \tag{5}$$

Third, if $y < u_1$, then by Fact 2, we have $y < u_1 \le s/3$ and $s - y - u_1 - \sqrt{yu_1} > 0$ which contradicts (5).

Thus, we have $y \ge u_1$. Also, when $x=z=(s-u_1)/2$ and $y = u_1$, we have $2y= 2u_1$.

The completes the proof of the claim.

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Proof of Fact 1:

“only if part”: Using $xyz \le (x + y + z)^3/27$, we have $27A^2 \le s^4$.

“if part”: Let us prove that there exist $x, y > 0$ such that $2x + y = s$ and $s x^2y = A^2$. Using $x = (s-y)/2$, we have $sy(s-y)^2 -4A^2 = 0$. Let $F(u) := sy(s-y)^2 - 4A^2$. We have $F(0) = -4A^2 < 0$ and $F(s/3) = \frac{4}{27}s^4 - 4A^2 \ge 0$. Thus, there exists $y_0 \in (0, s/3]$ such that $h(y_0) = 0$. Thus, $x = (s-y_0)/2, y = y_0$ satisfy $x, y > 0$, $2x + y = s$ and $sx^2y = A^2$.

Thus, $y = y_0, x = z = (s - y_0)/2$ satisfy $x,y,z>0$, $x + y + z = s$ and $(x + y + z)xyz = A^2$.

We are done.

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Proof of Fact 2:

Let $h(u) := su(s-u)^2 - 4A^2$.

If $27A^2 = s^4$, then $h(u) = -\frac{1}{27}s(4s-3u)(s-3u)^2$. Thus, $h(u) = 0$ has three positive real roots $s/3, s/3, 4s/3$.

If $27A^2 < s^4$, noting that $f(0) = -4A^2 < 0, f(s/3) = 4s^4/27 - 4A^2 > 0, f(s) = -4A^2 < 0$, $h(u) = 0$ has three positive real roots $0 < u_1 < s/3 < u_2 < s < u_3$.

We are done.