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For real numbers $a, b, c$ that satisfy $a + b + c = 6$ and $0\leq a,b,c \leq 4$, maximize $P=a^2+b^2+c^2+ab+ac+bc$.

My try:

$$\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$$ $$(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^2+bc=(6-a)^2-bc \end{cases} \end{align}$$ $$P=a^2+(b^2+c^2+bc)+a(b+c)=a^2+[(6-a)^2-bc]+a(6-a)$$ $$P=(a^2-12a+36)-bc=(a-6)^2-bc (2)\Rightarrow bc\leq 0 \Rightarrow P\geq (a-6)^2$$ When $bc=0 \Rightarrow [{\begin{matrix}b=0\\c=0\end{matrix}}(3)$. From $(1)$ and $(3)$, $\Rightarrow 2\leq a\leq 4(4)$ $P_{max} \implies |a-6|$ max satisfy $(4)$ $\implies a=2$ from $(1)$ and $(3)$ $\implies b=c=4$ $\implies P_{max}(a,b,c)=P(4;2;0)=28$

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    Hint: $P=a^2+b^2+c^2+ab+ac+bc=((a+b)^2+(b+c)^2+(a+c)^2)/2$ – 2017-02-15
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    ok thank you, this is a good suggest – 2017-02-15
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    Other hint : $P=\frac{1}{2}((a+b+c)^2+a^2+b^2+c^2)$ – 2017-02-15
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    note that the Maximum is $36$ – 2017-02-15
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    what? maximum is 36 at a=?; b=? ; c=? – 2017-02-15
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    @Dr.SonnhardGraubner The maximum is $28$. See my answer. – 2017-02-15
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    the Maximum is $36$ for $a=0,b=3,c=3$$ – 2017-02-15
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    @Dr.SonnhardGraubner: $3^2+3^2+3\cdot 3 = 27\neq 36$. – 2017-02-15
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    $a=0,b=3,c=3$ $Maximum=27$ – 2017-02-15
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    yes of Course $$9+9+9=27$$ – 2017-02-15
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    Why would users write 4 duplicate comments to one user? And @S.B.C, it's rather ironic that you call the Dr. on an error, when your comment addressing the Dr. is erroneous? This isn't tackle (American) football, where we have users very willing to form a mob to attack what's nothing more than a silly oversight. Yes, it is good to suggest that an error exists, and why. But we don't need three, four, five or more users to each separately attack. – 2017-02-15

3 Answers 3

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From the fact that $a+b+c=6$, we know that $$36=\left( \sum_{cyc} a \right)^2=\sum_{cyc}a^2+2 \sum_{cyc}ab=P+\sum_{cyc}ab$$

So maximizing $P$ becomes equivalent to minimizing $ab+bc+ca$.

WLOG, assume that $a$ is the maximum among $a,b,c$. So $a+b+c =6 \le 3a$, hence $4 \ge a \ge 2$ from the condition.

However, note that $$ab+bc+ca \ge ab+ca= a(b+c)=a(6-a) \ge 8$$ From $4 \ge a \ge 2$.

Hence the minimum of $ab+bc+ca=8$, when $a=4,b=2, c=0$. Thus, the maximum of $P$ is $36-8=28$.

We are done!

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    Why the downvotes? – 2017-02-15
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The stationary points of a quadratic form over a triangle (or hexagon) are simple to locate through Lagrange's multipliers. Your quadratic form is positive definite with eigenvectors $(1,1,1)$, $(-1,0,1)$, $(-1,1,0)$ hence the maximum value is attained on the boundary of the given domain, by convexity. By analyzing the instances $a=0$ and $a=4$ we easily get that the maximum is $28$, attained at $\{a,b,c\}=\{0,2,4\}$.

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    If $a=4, b=2, c=0$. Note that $P=4^2+2^2+4 \times 2=28$. – 2017-02-15
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    @S.C.B.: Indeed, Lagrange multipliers locate the minimum, and by convexity the maximum lies on the boundary. Thanks, now fixed. – 2017-02-15
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    I would upvote if I knew what you were doing. But you're welcome. – 2017-02-15
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    and since P is symmetric in $a,b,c$, and so are the bounds, then the same max is attained in every point with coordinates being a permutation of $(0,2,4)$ – 2017-02-15
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    Could someone explain why the downvote on this answer :(. Plese say why it is incorrect if you believe it to be so. – 2017-02-15
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Let $f(x)=x^2$ and $a\geq b\geq c$.

Hence, $f$ is a convex function and $(4,2,0)\succ(a,b,c)$.

Thus, by Karamata $$\sum_{cyc}(a^2+ab)=a^2+b^2+c^2+\frac{36-a^2-b^2-c^2}{2}=\frac{1}{2}(a^2+b^2+c^2)+18\leq$$ $$\leq\frac{1}{2}(4^2+2^2+0^2)+18=28.$$ Done!