Harder 3 unit.. (1 Viewer)

[gamecw]

man i can do all other 4u topics.. but doin badly in harder 3 unit..

especialy with inequality, i cant even do the basic ones..

can sumone solve these for me?

1.show that [a^2+b^2+c^2] is biger n = to [3 . 3rt of (a^2 . b^2 . c^2)]
2.show that [(a+b+c+d)/4] is biger n = to [4rt of abcd]
3. [x^3 +y^3+z^3] is biger n equal to 3xyz

they r questions r from New senior maths, 37(b)

 

[pLuvia]

2.
Considering the AM>GM property
(a+b)/2>sqrt{ab}
(c+d)/2>sqrt{cd}
Add these together
(a+b+c+d)/2>sqrt{ab}+sqrt{cd}
(a+b+c+d)/4>[sqrt{ab}+sqrt{cd}]/2
And since [sqrt{ab}+sqrt{cd}]/2>sqrt{sqrt{ab}{cd}}
Then
(a+b+c+d)/4>⁴rt{abcd}

 

[zeek]

3.

Always remember when doing stuff like this to look for patterns and to ALWAYS ALWAYS work with the inequalities a² + b²>=2ab and (a+b)²>=0

x² + y² >=2xy
.: x² + y² - xy>= xy

(x³ + y³)>=(x+y)(x²+y² -xy)
>=(x+y)xy
>=(x/z + y/z)xyz
Simliarly,

(z³ + y³)>=(z/x + y/x)xyz
(x³ + z³)>=(x/y + z/y)xyz

Adding all of these together...

2(x³ + y³ + z³) >=xyz[(x/y+y/x) + (y/z+z/y) + (z/x+x/z)]

Now before we proceed we need to simplify something for a/b + b/a...

a² + b² >= 2ab {let a=sqrt(a/b) and b=sqrt(b/a)}
.:a/b+b/a>=2

Returning...

2(x³ + y³ + z³) >=xyz[2+2+2]
2(x³ + y³ + z³) >=6xyz
.:x³ + y³ + z³ >=3xyz

 

[BIRUNI]

first one:
(a-b)^2=a^2-2ab+b^2>=0
a^2+b^2>=2ab
similarly
b^2+c^2>=2bc
a^2+c^2>=2ac

by adding these:
a^2+b^2+c^2>=ab+ac+bc
(a+b+c)^2-2ab-2ac-2bc>=ab+ac+bc
(a+b+c)^2>=3(ab+ac+bc)
can you do the rest now?

 

[zeek]

1.

Spoiler

Using the result from question 3...


x³ + y³ + z³ >=3xyz {Let x=a^2/3, y=b^2/3, z=c^2/3}

a²+b²+c²>=3(abc)^2/3
.: a²+b²+c²>=3³sqrt((abc)²)

EDIT: added spolier