proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$

Question

If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$

and also show that equality hold if $a=b=c$.

$\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$

Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$

So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$

Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$

$$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$

So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$

But I did not understand how can I prove above inequality

help Required

Thanks

Answer 1

For a triangle

$\Delta = \frac{abc}{4R} = rs$

Now in your inequality you can put in the values to get

$R \ge 2r$

This is known to be true since the distance between incentre and circumcentre $d^2 = R(R-2r)$

Thus your inequality is proved

Answer 2

It helps to write down what you want to prove:

$$(b+c-a)(c+a-b)(a+b-c) \le abc$$

You viewed the left-hand expression as a geometric mean, but replacing it by a arithmetic mean does not work because the resulting inequality is an AM-GM inequality in the wrong direction, so you need to try something else.

There are many ways to proceed, but a good routine first step is a substitution that replaces $a, b, c$ who are linked by triangle inequalities with $2x=b+c-a$, $2y=c+a-b$, $2z=a+b-c$ who are simply positive numbers.

The desired inequality becomes:

$$8xyz \le (y+z)(x+z)(x+y).$$

But this is simply a product of three AM-GM inequalities:

$$\sqrt{yz}\sqrt{xz}\sqrt{xy}\le \frac{y+z}2\frac{x+z}2\frac{x+y}2.$$

So everything is proved.

Alternatively, if you do not see it, you can simply expand the product and apply one AM-GM inequality to the 8 terms.

Answer 3

Thanks mathlove,phira,and user for solution.

My prove for $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$, where $a,b,c$ are the sides of a $\triangle$.

Using $\;\;\;\; \{b+(c-a)\}\cdot \{b-(c-a)\} = b^2-(c-a)^2\leq b^2$

similarly $ \{c+(a-b)\}\cdot \{c-(a-b)\} = c^2-(a-b)^2\leq c^2$

similarly $\{a+(b-c)\}\cdot \{a-(b-c)\} = a^2-(b-c)^2\leq a^2$

Multimply these three equations, we get $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$

and equality hold when $a=b=c$

Answer 4

We know $$S=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}.$$

So, all we need is to prove the following : $$(-a+b+c)(a-b+c)(a+b-c)\le abc.$$

This is a well known inequality. Proof is here.