# Inequality equation of a Triangle

I have divided the Problem into two parts, a) and b):

a) Let $$a$$, $$b$$ and $$c$$ be the side lengths of a triangle with a perimeter of $$4$$. I need to prove that

$$a^2 + b^2 + c^2 + abc < 8$$.

b) And is there a real number $$d<8$$ such that for each triangle with the perimeter $$4$$, the inequality equation

$$a^2 + b^2 + c^2 + abc

is valid?

I tried to use Heron's Formula somehow for the first inequality equation.

$$s \, = \, \frac{a+b+c}{2}$$

and

$$F_{\triangle} = \sqrt{s(s-a)(s-b)(s-c)}$$

Now we have:

$$2s=4$$ so that $$s=2$$

Thus:

$$F_{\triangle} = \sqrt{2(2-a)(2-b)(2-c)}$$

Maybe we can exchange $$F_{\triangle}$$ somehow?

Has anyone another approach or an idea how to continue with Heron's formula?

Thx

• what do you mean by circumference of a triangle? Are you referring to the perimeter? – Anurag A Nov 18 '18 at 18:27
• Yes, I'll change it... – calculatormathematical Nov 18 '18 at 18:28

a) Because of $$a+b+c=4$$, you have $${a^2}+{b^2}+{c^2}={(a+b+c)^2}-2(ab+bc+ac)=4(a+b+c)-2(ab+bc+ac)$$ Therefore $${a^2}+{b^2}+{c^2}+abc=4(a+b+c)-2(ab+bc+ac)+abc=8-(2-a)(2-b)(2-c)$$
By triangle-inequality $$b+c>a$$ $$\Rightarrow4=a+b+c>a+a=2a\Rightarrow2>a$$ Analugously you can prove that $$2>b$$ and $$2>c$$.
Thus $${a^2}+{b^2}+{c^2}+abc=8-(2-a)(2-b)(2-c)<8$$
b) This inequality for $$d<8$$ isn't necessarily valid. Let for instance $$k=1-\frac{d}{8}>0$$.
Let furthermore $$a=b=2-k>0$$ Hence $${a^2}+{b^2}+{c^2}+abc>{a^2}+{b^2}=2{(2-k)^2}=8-8k+{k^2}>8-8k=d$$ which contradicts the condition $${a^2}+{b^2}+{c^2}+abc