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Working on a problem...

Find the least positive real number $k$ such that $7\sqrt{a} + 17\sqrt{b} + k\sqrt{c} \ge \sqrt{2019}$ over all positive real numbers $a,b,c$ with $a+b+c=1$.

Maximizing the "$a$" term doesn't seem to work, and expansion through moving a radical to the right side of the equation simply leads to more radicals.

Any help?

cqfd
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Hubert Lioni
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  • I would just write $k=\frac {\sqrt{2019}-7\sqrt a-17\sqrt b}{\sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations. – Ross Millikan Jan 10 '19 at 06:11

2 Answers2

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This response is informal handwaving, which I suspect leads to the accurate answer. Any value apportioned to a, or b, is deducted from c.

Since $\sqrt{2019} > 17, k$ must be greater than 17. Therefore the solution lies in maximizing c and setting k correspondingly. With a=0=b, and c=1, then $k=\sqrt{2019}.$

I will be very surprised if this informal approach is wrong.

Addendum

It occurs to me that I may have misinterpreted the problem. If it is desired to identify K so that the inequality holds regardless of how a,b,c are apportioned (as long as a,b,c positive, a+b+c=1) then there is no answer. Regardless of the fixed value of k, as c approaches 0, $k\sqrt{c}$ will approach 0, and the left hand side will then be no greater than 17.

Addendum 2

Are you sure that you have the constraints right?

user2661923
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It cannot be done. Put $$a=1-2\epsilon^2,\quad b=c=\epsilon^2$$ with $0<\epsilon\ll1$. Then we want $$7\sqrt{1-2\epsilon^2}+(k+17)\epsilon\geq\sqrt{2019}\ ,$$ or $$(k+17)\epsilon\geq44.933-7\sqrt{1-2\epsilon^2}>37\ .$$ There is no $k$ that can do this for all $\epsilon>0$.

Christian Blatter
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