Problem in mathematics from Sequence and Series..plz help me..?

Question

Prove that for positive numbers a and b, (a+b)(1/a+1/b)>= 4

Prove that (xyz)^1/3.(1/x+1/y+1/z)>= 3 where x,y,z belongs to real positive numbers

I m really getting frustrated cuz i m nt able to solve these 2 problems.. it is from Sequence and Series chapter.. I m trying to be calm ...I just need solution of these 2 question..Thanks.. any one good in mathematics plz leave ur id plz plz plzz..
Thanks in advance

Answer 1

Since (a+b)*(1/a + 1/b) can be written as
(a+b)/a + (a+b)/b then the total is
1 + b/a + 1 + a/b = 2 + a/b + b/a = 2 + (a² + b²)/ab

It should be pretty obvious that the smallest value for this function is when a = b and that's 4. Now if a = kb (where k is some constant k>1) then
(a² + b²)/ab = (k²b² + b²)/kb² = (k²+1)/k which will be > 2 for all k > 1

Now **you** do the second one so you actually learn something.


Doug

Answer 2

For the first,

(a+b).(1/a + 1/b) = 2 + (b/a) + (a/b)
this is evidently greater than 2, but you ought to notice how these fractions appear,
for a,b E Z+
as b gets smaller b/a decreases but a/b increases and it's the same with a.
So a=b should be the case which yields the lowest value.
then,
(a+b)/(1/a + 1/b) >= 4
the second,
(xyz)^(1/3) . (1/x + 1/y + 1/z) = (xyz)^(1/3) . [(x+y+z)/xyz]

= [(xyz)^ (1/3 - 1)].[x+y+z]
= [(x+y+z)/ (xyz)^(2/3)]

Now all you need to note is that x,y,z E Z+
Try it yourself...

Answer 3

a tennis journey is held annuly. contained interior the 1st around sixty 4 suits are complete. in each successive around the form of suits complete decreases by ability of potential of a million/2. answer: by ability of potential of undertaking-loose journey you will see that the form of suits for the rounds are sixty 4 32 sixteen 8 4 2 a million yet you will now no longer write this notation. it would want to be like this: B = sixty 4(a million/2)^(n - a million) the area B = form of suits for the nth around you notice that there is not any 0'th around or a million/2 form of suits, so which you decrease: 0 < n < 8 and n is an integer the total is: ?(ok = a million to 7) sixty 4(a million/2)^(ok - a million) Use the formula ?(ok = a million to n) ar^(ok - a million) = a(a million - r^n)/(a million - r) subsequently, ? = sixty 4(a million - a million/2 ^ 7)/(a million - a million/2) ? = sixty 4(a million - a million/128)/(a million - a million/2) ? = sixty 4(127/128)/(a million/2) ? = 127 ^_^

Answer 4

(a-b)^2 >= 0
=>a^2+b^2-2ab >= 0
=> a^2 + b^2 >= 2ab
=> a^2+b^2+2ab >.= 2ab+2ab
=> (a+b)^2 >= 4ab
=> (a+b)(a+b) >= 4ab
=>. (a+b)(a+b)/ab >= 4
=> (a+b)(1/a+1/b0 >= 4