Proof inequality q (1 Viewer)

Run hard@thehsc · Feb 2, 2022

I investigated this question and worked backwards to front, but I am not sure if this is right - could you guys check it pls!!!

5uckerberg · Feb 2, 2022

Okay, @Run hard@thehsc can you show us how you answered this question? We might be of help.

5uckerberg · Feb 2, 2022

Okay, in that case, we will start. $\left(\sqrt{x}-\sqrt{y}\right)^{2}\geq{0}$ and then the next step $\frac{x+y}{2}\geq{\sqrt{xy}}$ . This is done by expanding and bringing the square root term on the other side.

Well, here we can let $x=\frac{a+b}{2}, y=\frac{c+d}{2}$ .
After subbing the values of x and y we will have $\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2} \geq{ \sqrt{\sqrt{ab}\sqrt{cd}}}$ .
Further simplifications lead us to $\frac{a+b+c+d}{4}\geq{\sqrt[4]{abcd}}$ .
Now bon appetit

=)(= · Feb 2, 2022

5uckerberg said:
Okay, in that case, we will start. $\left(\sqrt{x}-\sqrt{y}\right)^{2}\geq{0}$ and then the next step $\frac{x+y}{2}\geq{\sqrt{xy}}$ . This is done by expanding and bringing the square root term on the other side.

Well, here we can let $x=\frac{a+b}{2}, y=\frac{c+d}{2}$ .
After subbing the values of x and y we will have $\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2} \geq{ \sqrt{\sqrt{ab}\sqrt{cd}}}$ .
Further simplifications lead us to $\frac{a+b+c+d}{4}\geq{\sqrt[4]{abcd}}$ .
Now bon appetit

A bit lost as to how you got the rhs of the inequality in this line

as i cant see how you subsituted x=a+b/2 and y=c+d/2?

5uckerberg · Feb 2, 2022

Well simply put recall that $\frac{a+b}{2}\geq{\sqrt{ab}}, \frac{c+d}{2}\geq{\sqrt{cd}}$ and with these two replace the x and y with them.

=)(= · Feb 2, 2022

5uckerberg said:
Well simply put recall that $\frac{a+b}{2}\geq{\sqrt{ab}}, \frac{c+d}{2}\geq{\sqrt{cd}}$ and with these two replace the x and y with them.

Got it thank you

Run hard@thehsc · Feb 7, 2022

@5uckerberg sos for the late reply - but it makes sense...thank you!

Proof inequality q (1 Viewer)

Run hard@thehsc

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5uckerberg

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5uckerberg

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5uckerberg

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Run hard@thehsc

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