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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

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

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

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 $b+c>a$ $$Rightarrow4=a+b+c>a+a=2aRightarrow2>a$$
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 $${a^2}+{b^2}+{c^2}+abc<d$$