MHB Inequality Proof: x/sqrt(2y^2+5) + y/sqrt(2x^2+5) <= 2/sqrt(7)

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    2016
AI Thread Summary
The discussion presents a mathematical inequality to prove: (x/sqrt(2y^2+5)) + (y/sqrt(2x^2+5)) ≤ 2/sqrt(7) for 0 ≤ x, y ≤ 1. Participants are encouraged to submit their solutions following the guidelines provided in the forum. Acknowledgment is given to kaliprasad for successfully solving the problem. The thread emphasizes the importance of collaborative problem-solving in mathematics. Engaging with such inequalities enhances understanding of mathematical concepts and techniques.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----
Let $0\le x \le 1$ and $0\le y \le 1$ . Prove the inequality

$$\frac{x}{\sqrt{2y^2+5}}+\frac{y}{\sqrt{2x^2+5}}\le \frac{2}{\sqrt{7}}.$$

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
Congratulations to kaliprasad for his correct solution::)

You can find the proposed solution as follows:
Since $x^2\le 1$, we have $2y^2+5\ge 2y^2+2x^2+3.$

Similarly, we have $2x^2+5\ge 2x^2+2y^2+3$ and it follows that

$$\frac{x}{\sqrt{2y^2+5}}+\frac{y}{\sqrt{2x^2+5}}\le \frac{x+y}{\sqrt{2y^2+2x^2+3}}$$

Therefore, it suffices to show that $\sqrt{7}(x+y)\le 2\sqrt{2y^2+2x^2+3}$. Squaring and rearranging the terms, this is equivalent to $12xy\le (x-y)^2+12$, which is certainly true since $12xy\le 12$ and $(x-y)^2+12\ge 12$.

The result follows. Equality holds if and only if $x=y=1.$
 
Back
Top