I think it's inspired by some posts of mine from here and my forums,since July 2017. But it's beautiful.Let's mention that RavaloD was the solver back then. But since the mathematics is tricky,can be a simple coincidence.

Tintarn wrote: a) Let $a,b$ and $c$ be side lengths of a triangle with perimeter $4$. Show that \[a^2+b^2+c^2+abc<8.\]b) Is there a real number $d<8$ such that for all triangles with perimeter $4$ we have \[a^2+b^2+c^2+abc<d \quad\]where $a,b$ and $c$ are the side lengths of the triangle?

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The problem is an application of the following equality for any triangle: $ (a+b+c)^2+2r(R-2r)=2(a^2+b^2+c^2)+\frac{9abc}{a+b+c} \ \ ; $ From here it is easy to see why they asked for 8, as $max$. $ 16+2r(R-2r)=2(a^2+b^2+c^2)+ \frac{9abc}{4} \ \ ; $ $ 8+r(R-2r)= a^2+b^2+c^2+ abc+ \frac{abc}{8} \ \ ; $ Now you see that $\frac{abc}{8}=\frac{4Rsr}{4s}=Rr$ and $ Rr-Rr+2r^2 > 0 \ \ ; $ and of course: $ 8-2r^2=a^2+b^2+c^2+abc \ \ ; $ $ 8 >a^2+b^2+c^2+abc \ \ ; $ Greetings from Lorian Saceanu

Tintarn wrote: a) Let $a,b$ and $c$ be side lengths of a triangle with perimeter $4$. Show that

\[a^2+b^2+c^2+abc<8.\]*

The inequality is equivalent to $$(b+c-a)(c+a-b)(a+b-c)>0.$$

Now the general problem.

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If $a, b, c $ are the sides of $\triangle ABC$ and $a+b+c=1$, then: $ 5(a^2+b^2+c^2)+18abc \ge \frac{7}{3} \ \ ; $ From a book! Greetings!

Not bad,but this is a minimum problem.Which is not so complicated as the maximum problem posted by me. That problem covers ALL.

Tintarn wrote: b) Is there a real number $d<8$ such that for all triangles with perimeter $4$ we have \[a^2+b^2+c^2+abc<d \quad\]where $a,b$ and $c$ are the side lengths of the triangle? There isn't. Put $a=x+y,b=y+z,c=z+x$. If $x=0$, we have \[a^2+b^2+c^2+abc=8.\]So take $x\rightarrow 0^+$ and $[a^2+b^2+c^2+abc\rightarrow 8$.

mihaig wrote: I think it's inspired by some posts of mine from here and my forums,since July 2017. But it's beautiful.Let's mention that RavaloD was the solver back then. But since the mathematics is tricky,can be a simple coincidence. Maybe, but I don't think so. Hungary 1990 Let $a,b$ and $c$ be side lengths of a triangle with perimeter $2$. Show that \[a^2+b^2+c^2+2abc<4.\]Bonus from me Is there a real number $d<4$ such that for all triangles with perimeter $2$ we have \[a^2+b^2+c^2+2abc<d \quad\]where $a,b$ and $c$ are the side lengths of the triangle?

See here bonus.It's waiting from last year.

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$(b+c-a)(c+a-b)(a+b-c)>0\implies\max (a^2+b^2+c^2+kabc)\le 27k+\left(1-\frac{3}{2}k\right)\sum a^2$ But I guess I'm asked to write $\max (a^2+b^2+c^2+kabc)=f(k)$ right?

WolfusA wrote: $(b+c-a)(c+a-b)(a+b-c)>0\implies\max (a^2+b^2+c^2+kabc)\le 27k+\left(1-\frac{3}{2}k\right)\sum a^2$ But I guess I'm asked to write $\max (a^2+b^2+c^2+kabc)=f(k)$ right? Of course,there is a good reason for which is unsolved yet. Because it's difficult,and should not be treated superficially .

For part a), use Ravi substitution and expand to arrive at $$2\sum_{cyc}x^2 + 2\sum_{cyc}xy+\sum_{sym}x^2y+2xyz < 8$$. Now homogenize using $x+y+z=2$ and one ends up with $$\sum_{cyc}x^3+3\sum_{sym}x^2y+5xyz < \sum_{cyc}x^3+3\sum_{sym}x^2y+6xyz$$i.e. $xyz > 0$, which is trivially true. For b), just take $a = \sqrt{\frac{d}{2}} + \epsilon, b = \sqrt{\frac{d}{2}} +2 \epsilon, c=2\epsilon$ where $\epsilon \to 0$ (and $a+b+c=4$) to get a contradiction for any $d < 8$.

mihaig wrote: Now the general problem. $$max\left \{ a^2+b^2+c^2+kabc \right \}=\left\{\begin{matrix} 18 &, 0<k\leq \frac{2}{3} & \\ &&\\ \frac{27}{4}\left ( k+2 \right ) &,\frac{2}{3}<k\leq \frac{6}{5} & \\ &&\\ 8k+12 &, k>\frac{6}{5} & \end{matrix}\right.$$

anhduy98 wrote: mihaig wrote: Now the general problem. $$max\left \{ a^2+b^2+c^2+kabc \right \}=\left\{\begin{matrix} 18 &, 0<k\leq \frac{2}{3} & \\ &&\\ \frac{27}{4}\left ( k+2 \right ) &,\frac{2}{3}<k\leq \frac{6}{5} & \\ &&\\ 8k+12 &, k>\frac{6}{5} & \end{matrix}\right.$$ It's the best answer I've got. But without a rigurous proof I can't figure if it's well or not. Thank you for your interest.

for 2/3<k<=1 I get that the maximum is achieved when a=b=2/k. Other answers are like mine.

a) This was also something similar Bosnia JBMO TST 2012.. And the idea was to note that $a+b>c$ hence $4>2c$ or $(2-a)(2-b)(2-c)>0$ from where the claim is evident.

A slightly different approach using Ravi Substitution: Set $a = x+y, b = y+z, c = z+x$ with $x+y+z = 2$. Then \begin{align*} a^2+b^2+c^2+abc &<8 \\ \iff \sum_{\operatorname{cyc}} (x+y)^2 + (x+y)(y+z)(z+x) &< 8 \\ \iff 2\sum_{\operatorname{cyc}} x^2 + 2\sum_{\operatorname{cyc}} xy + \sum_{\operatorname{sym}} x^2 y + 2xyz &< 8 \\ \iff \sum_{\operatorname{cyc}} x^2 + (x+y+z)^2 + \sum_{\operatorname{sym}} x^2 y + 2xyz &<8 \\ \iff \sum_{\operatorname{cyc}} x^2 + \sum_{\operatorname{sym}} x^2 y + 2xyz &< 4 \\ \iff \sum_{\operatorname{cyc}} x^2 + \sum_{\operatorname{cyc}} xy(x+y+z) -xyz &<4 \\ \iff (x+y+z)^2 - xyz &<4 \\ \iff -xyz &< 0. \end{align*}Let $x \to 0$ to show that $8$ is the best constant.

I got a different approach that should work as well. Consider the function $f(a)=a^2+abc+(b^2+c^2)$ where $a\in[0,2]$. We choose this interval as none of the variables could be bigger than 2 by the triangle inequality. As $f$ is convex in each of these variables the maximum will occur only if $a,b,c\in\{0,2\}$. The triples $(a,b,c)=(2,2,0)$ and it's permutations is are now the triples that satisfy the problem condition, so $a^2+b^2+c^2+abc\le 8$. However, as they are sides of a triangle, none of them can be 0, so the inequality is strict. b) is trivial by just pushing on variable to the endpoints.