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If $ 4a^2+3ab+b^2\le2016 $, then what can we say about the maximum value of $a+b$? Here $a, b$ are real numbers.

I think we may use Cauchy-Schwarz inequality here. Any ideas thanks.

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    This is problem [J395](https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2016-06/mr_6_2016_problems_1.pdf) with a January 15, 2017 deadline. I am asking the moderators to lock this problem until then. – 2017-01-13

3 Answers 3

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Firstly, $$4a^2+3ab+b^2=(2a+b)^2-ab\le2016$$ or equivalently $$(2a+b)^2\le2016+ab$$ Now you have, from AM-GM inequality, $$\frac{(a+b)^2}{2}\le \frac{(2a+b)^2}{2}\le1008+\frac{ab}{2}\le 1008 +\frac{(a+b)^2}{4}$$ Then you have from first and last part of inequality $$\frac{(a+b)^2}{4} \le 1008$$ or $$a+b \le \sqrt {4032}$$

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    superb answer. Easy one though! – 2017-01-03
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    @vidyarthi The answer is incorrect, for example we can have $a=-12$ and $b=60$ that satisfies $4a^2+3ab+b^2=2016$ but $a+b=48 > \sqrt{2016}$. – 2017-01-03
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    @AnuragA modified the post. – 2017-01-03
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    even $\sqrt{4032}$ is way too high as an estimate for the maximum value. In fact there exist no $a,b$ for which it will attain that bound. – 2017-01-03
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    @AnuragA what bound do you suggest and how? – 2017-01-03
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    @vidyarthi windircurse is assuming that $a,b >0$, which is incorrect and hence the usage of AM-Gm is invalid. – 2017-01-03
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    @AnuragA thanks, so then is Cauchy-Schwarz the way? – 2017-01-03
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You are trying to maximise a linear function, in a nice convex region. This will happen when the line $a+b = c$ is tangent to the boundary $C: 4a^2+3ab+b^2=2016$.

The tangent is given by $C' = 0 \implies 8a+ 3ab'+3b + 2bb'=0 \implies b' = -\dfrac{8a+3b}{3a+2b} = -1 \implies b = -5a$. Using this in the curve, the tangent points are when $a=\pm 12, b = \mp 60$, so the maximum is when $a=-12, b = 60 \implies a+b = 48$.

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    how is $b'=-1$? – 2017-01-10
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    @vidyarthi that's the slope of any line $a+b=c$ and so we look for points on the curve where it matches. – 2017-01-10
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If you want a method without calculus and with Cauchy Schwarz, we first transform the ellipse to a simpler form: $$2016\times 4= 4(4a^2+3ab+b^2) = 7a^2 + (3a+2b)^2$$ Now, using CS inequality, $$(7a^2+(3a+2b)^2)(\tfrac17+1) \geqslant (-a + 3a+2b)^2 = 4(a+b)^2$$

$$\implies 2016\times \frac87 \geqslant (a+b)^2 \implies a+b \leqslant 48$$ As $(a, b) = (-12, 60)$ achieves equality, this is indeed the maximum.