The Answer Is Not 0.5

Algebra Level 5

What is the smallest real number $k$ (to 3 decimal places), such that for all ordered triples of non-negative reals $(a,b,c)$ which satisfy $a + b + c = 1$ , we have

$\frac{ a^2 + bc} { a+1} + \frac{ b^2+ca} { b+1} + \frac{ c^2+ab} { c+1} \leq k?$

If you want a similar inequality, you can try The Answer Is Not 0.225 .

The answer is 0.583.

1 solution

Calvin Lin Staff
Apr 11, 2014

(This is not a solution.)

The maximum is achieved at $(0, \frac{1}{2}, \frac{1}{2} )$ and cyclic permutations, which yield $k = 0.5833$ .

Now it's up to you to prove that

$\frac{ a^2 + bc} { a+1} + \frac{ b^2+ca} { b+1} + \frac{ c^2+ab} { c+1} \leq \frac{7}{12}.$

Even in this question (and the previous one) and in the problem 'Maximizing A cyclic polynomial', I saw that the maxima was attained when one of the variables was zero. Is it always true that if the maxima of symmetric constrained functions of this type is not attained when all the variables are equal, then it will be attained when one of them is zero. Do you know a question in which the case is not so?

Led Tasso - 7 years, 1 month ago

These functions were 'specially' chosen, and hence have 'nice' solution sets.

It is not necessary that the extrema must occur on a boundary (in this case, $x = 0$ ), or when some/all of the variables are equal. For example, consider the following:

If $x , y$ are non-negative real numbers such that $x+ y = 1$ , what is the minimum of $[ (x - \frac{1}{4} ) ( y - \frac{1}{4} ) ] ^2$

In this case, the minimum is clearly 0, and occurs when either $x = \frac{1}{4}$ or $y = \frac{1}{4}$ , neither of which are boundary conditions.

Calvin Lin Staff - 7 years, 1 month ago

Thanks for that.

Led Tasso - 7 years, 1 month ago

Why isn't maximum at $a=b=c=\frac{1}{3}$ ? Here's my logic. Consider an inequality of root mean square and arithmetic mean, hence we obtain: $\sqrt{\frac{\sum_{cyc}(\frac{a^{2}+bc}{a+1})^{2} }{3}}\geq \frac{\sum_{cyc}(\frac{a^{2}+bc}{a+1})}{3}$ Then, we account for the restriction of $a+b+c=1$ . So we have, $\sqrt{\frac{\sum_{cyc}(\frac{a^{2}+bc}{2a+b+c})^{2} }{3}}\geq \frac{\sum_{cyc}(\frac{a^{2}+bc}{2a+b+c})}{3}$ With that, the inequality is suppose to be in equality at $a=b=c$ . Hence, $3\sqrt{\frac{(\frac{2a^{2}}{4a})^{2}\times 3}{3}}=\sum_{cyc}(\frac{a^{2}+bc}{2a+b+c})$ $\frac{3a}{2}=\sum_{cyc}(\frac{a^{2}+bc}{2a+b+c})$ Since $a=b=c$ and $a+b+c=1$ , we get $a=\frac{1}{3}$ . Finally, we plug in $a$ into the summation to get: $\sum_{cyc}(\frac{a^{2}+bc}{2a+b+c})=\boxed{\frac{1}{2}}$ But apparently this isn't the answer. What's wrong with this solution?

Shaun Loong - 6 years, 11 months ago

It is a very common mistake to assume that the maximum / minimum must occur when all the variables are equal.

You have not shown that $\sqrt{ \frac{ \sum ( \frac{ a^2+bc} {2a+b+c}) ^2 } { 3} }$ is indeed bounded above by $\frac{1}{2}$ . All that you have shown, is that the value is equal to $\frac{ 1}{2}$ if $a=b=c$ . That doesn't tell you what the value of the function is at other points.

Calvin Lin Staff - 6 years, 10 months ago

Hey, Sorry for bumping this old question but I still don't understand why 0.5833 should be the lowest when k can attain 0.5.Could you help?

Vishnu Bhagyanath - 6 years ago

@Vishnu Bhagyanath – $0.5 \leq 0.5833$ .

When $a , b, c = ( 0 , 0.5, 0.5)$ , then the LHS is equal to 0.5833. If $k = 0.5$ , then we have $0.5833 \not \leq 0.5$ .

Calvin Lin Staff - 6 years ago

How would you show that?

Chung Kevin - 6 years, 10 months ago

comment as-tu su que le maximum est atteint à (0,1/2,1/2)?

Omar El Mokhtar - 7 years, 2 months ago

Il veut que vous à le résoudre par vous-même. :D

Finn Hulse - 7 years, 1 month ago

The Answer Is Not 0.5

The answer is 0.583.

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