Trial Q (1 Viewer)

krypticlemonjuice · Sep 26, 2023

Can someone please do this question and show me their working?

I have no idea what this working out means, especially the first line

synthesisFR · Sep 26, 2023

i see these inequality questions, i skip

carrotsss · Sep 26, 2023

hard inequalities question

(a^3+1)(b^3+1)(c^3+1) \geq (\frac{ab+bc+ca+1}{2})^3 I found this question in the JRHS 2023 paper that was upload here recently. I skipped it when I did the practise test, but when I went over it again I still couldn't get it. I asked my teacher and she hasn't gotten it as of yet. Could anyone...

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Luukas.2 · Sep 26, 2023

krypticlemonjuice said:
View attachment 39983

Can someone please do this question and show me their working?

View attachment 39984

I have no idea what this working out means, especially the first line

In part (ii), you showed that

$a^3 + b^3 + c^3 \ge 3abc$

so long as a, b, and c are real and positive.

$\text{Let } a = \sqrt[3]{\cfrac{x^3}{1+x^3}} \qquad \text{and} \qquad b = \sqrt[3]{\cfrac{y^3}{1+y^3}} \qquad \text{and} \qquad c = \sqrt[3]{\cfrac{1}{1+z^3}}$

where x, y, and z are suitably chosen positive reals. Putting these into the equation from part (ii):

$\text{LHS(from part ii) } = a^3 + b^3 + c^3 = \cfrac{x^3}{1+x^3} + \cfrac{y^3}{1+y^3} + \cfrac{1}{1+z^3}$

$\text{RHS(from part ii) } = 3abc = 3\sqrt[3]{\cfrac{x^3}{1+x^3} \times \cfrac{y^3}{1+y^3} \times \cfrac{1}{1+z^3}} = \cfrac{3\sqrt[3]{x^3 y^3}}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}} = \cfrac{3xy}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}$

and thus you get the solution's inequation (1), except in x, y, and z.

Similar substitutions yield the other three inequations, and they sum to give the result required.

The problem with this approach is that the substitutions collectively require a, b, and c to be less than 1.

synthesisFR · Sep 26, 2023

i still cant comprehend what happened

unofficiallyred12 · Sep 28, 2023

Luukas.2 said:
In part (ii), you showed that

$a^3 + b^3 + c^3 \ge 3abc$

so long as a, b, and c are real and positive.

$\text{Let } a = \sqrt[3]{\cfrac{x^3}{1+x^3}} \qquad \text{and} \qquad b = \sqrt[3]{\cfrac{y^3}{1+y^3}} \qquad \text{and} \qquad c = \sqrt[3]{\cfrac{1}{1+z^3}}$

where x, y, and z are suitably chosen positive reals. Putting these into the equation from part (ii):

$\text{LHS(from part ii) } = a^3 + b^3 + c^3 = \cfrac{x^3}{1+x^3} + \cfrac{y^3}{1+y^3} + \cfrac{1}{1+z^3}$

$\text{RHS(from part ii) } = 3abc = 3\sqrt[3]{\cfrac{x^3}{1+x^3} \times \cfrac{y^3}{1+y^3} \times \cfrac{1}{1+z^3}} = \cfrac{3\sqrt[3]{x^3 y^3}}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}} = \cfrac{3xy}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}$

and thus you get the solution's inequation (1), except in x, y, and z.

Similar substitutions yield the other three inequations, and they sum to give the result required.

The problem with this approach is that the substitutions collectively require a, b, and c to be less than 1.

why is a, b, c being less than 1 a problem tho cos ii is true for all positive a, b, c

Trial Q (1 Viewer)

krypticlemonjuice

Member

synthesisFR

しゃけ

carrotsss

New Member

hard inequalities question

Luukas.2

Well-Known Member

synthesisFR

しゃけ

unofficiallyred12

Member

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