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Let $x$ and $y$ be real numbers, such that $x \ge 1$ and $y \ge 1$.

Prove that this inequality is true: $\sqrt{x-1} + \sqrt{y-1} \le xy$

Can someone show me steps to solve it. PS:I need to give steps on how to solve it.

hikaru jakafura
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    "logic"???............. – Mauro ALLEGRANZA May 20 '22 at 17:45
  • @MauroALLEGRANZA => yes, I have to solve it while showing the steps logically not just here is the solution. – hikaru jakafura May 20 '22 at 17:47
  • A mathematical logic course is a incredibly different beast, to explain Mauro's comment – FShrike May 20 '22 at 17:51
  • @FShrike yes we study a subject in it and we get some simple logic problems like here is this propositions show that they are true or false ... and prove that but one of those questions was this above which I have to show that I understand {conjunction and disjunction} by proving that the inequality above is true for all x >=1 and y>=1...

    PS: my english is bad so I can't explain here a lot in english I just try

     – hikaru jakafura May 20 '22 at 17:56
  • One way to show it is with calculus. This is probably a bit overkill, but do you know about differentiation and the identification of turning points / increasing properties? – FShrike May 20 '22 at 17:58

3 Answers3

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Both sides are positive. So it is an equivalent inequality to the one you get from squaring both sides.

Then do some algebra to isolate a square root again. It should look like:

$$2\sqrt{(x-1)(y-1)} ≤ x^2y^2 - x - y + 2$$

But since $x\geq1$, then $x^2\geq x$ and similar for $y$. So it suffices to show

$$2\sqrt{(x-1)(y-1)} ≤ xy - x - y + 2=(x-1)(y-1)+1$$

Now you have a situation where $c$ is some positive number and you need to show that $2\sqrt{c}\leq c+1$. Can you do this last step?

2'5 9'2
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  • isn't in the first inequality you wrote, there should be 2$\sqrt((x-1)(y-1))$ I think you forgot the 2 there, but I understand the idea of what you did if you can explain what to do with that 2 – hikaru jakafura May 21 '22 at 10:01
  • @hikarujakafura Thanks, good eye for detail. Keeping track of that $2$ doesn't chage things, it turns out. – 2'5 9'2 May 21 '22 at 15:56
  • @ 2'59'2 okk thank you for this solution, I want to ask you something I see a lot of people use (Am gm, ...) these techniques to solve inequalities , did you use any of those techniques here or how did you come to think of the solution , because sometimes I take days with one problem I can't solve it until someone gives me the solution – hikaru jakafura May 21 '22 at 17:21
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    @hikarujakafura When I see an equation or inequality with two radicals like this one, I consider arranging things so the two radicals are alone on one side, and then squaring will reduce to only having one radical. Repeat that and you can have no radicals. You have squared twice so the terms may be messy now. But sometimes it helps clear some things up. This is what first occurred to me to look into. – 2'5 9'2 May 22 '22 at 01:16
  • ok thank you so much – hikaru jakafura May 22 '22 at 12:47
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Using Cauchy Schwarz and AM-GM

$$\sqrt{x-1}+\sqrt{y-1} \le \sqrt{2(x+y-2)} \le \frac{2(x+y-2)+1}{2} \le x+y-1 $$

Now it suffices to prove that

$$x+y-1 \le xy$$

Wich is equivalent to $$(x-1)(y-1) \ge 0$$

So we are done.

ali3985
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you can use schwartz inequality directly

we know ;$a_1 b_1+a_2b_2 \leq \sqrt{a_{1}^2+a_{2}^2} \sqrt{b_{1}^2+b_{2}^2}$

just put $a_1 =\sqrt{x-1}$

$a_2 =1$

$b_1 =1$

and

$b_2 =\sqrt{y-1}$

so you will get ;$\sqrt{x-1} .1+1.\sqrt{y-1} \leq \sqrt{x-1+1}\sqrt{y-1+1}=\sqrt{xy}$