<s>[INST] 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
$a^2 + b^2 + c^2 + abc <d$
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
 [/INST] 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 $${a^2}+{b^2}+{c^2}+abc<d$$
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