Prove these inequalities?

a) a^4 + b^4 >/= a^3b + ab^3

b) a^2 + b^2 + c^2 >/= ab + bc + ac


ans:
a)
we have following from AM GM enaquality

a^4 + (ab)^2 >= 2a^3.b .. 1
b^4 + (ab)^2 >= 2b^3.a ... 2
a^4 + b^4 >= 2(ab)^2 .. 3

add to get 2(a^4+b^4 ) + 2a^2b^2 >= 2a^3b + 2 ab^3 + 2 a^2 b^2
or (a^4+b^4 ) >= a^3b + ab^3

hence proved
b) we know a^2 + b^2 > = 2ab (from AM GM enaquality)
b^2 + c^2 >= 2bc
c^2 + a^2 >= 2ac

add to get 2(a^2 + b^2 + c^2) >= 2(ab+bc+ca)
or (a^2 + b^2 + c^2) >= ab + bc + ca

2011/105) Prove a^4 + b^4 + c^4 >= abc(a+b+c)

this can be done in 2 steps

we know by AM GM inequality

a^4+b^4 > = 2 a^2b^2
b^4+c^4 >= 2 b^2 c^2
c^4 + a^4 >= 2 a^2^2

adding the 3 above and deviding by 2 we get

a^4+b^4+c ^4 >= a^2b^2+b^2c^2 + c^2 a^2 ...1

now we repeat the process of a^2b^2 , b^2 c^2 and c^2 a^2 to get as below

a^2 b^2 + b^2 c^2 > = 2 b^2ac
b^2c^2 + c^2 a^2 >= 2 c^2ab
c^2a^2 + a^2 b^2 >= 2 a^2bc

adding the above and deviding by 2 we get

a^2b^2 + b^2 c^2 + c^2 a^2 >= (b^2ac+c^2ab+a^2bc) or abc(b+c+a) ,.2

from (1) and (2) it follows

a^4 + b^4 + c^4 >= abc(a+b+c)

2011/101) Prove this inequality? 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x (x is a positive number)

we shall use the AM GM enaquality to prove it

we know (a+b)/2 > = sqrt(ab) or (a+b) > = 2 sqrt(ab)

a = 4x^10, b = x^8 gives 4x^10 + x^ 8 > = 4 x^9
a = 4x^10, b = 1 gives 4x^10 + 1 > = 4 x^5
a = x^8, b = 4x^2 gives x^ 8 +4x^2 > = 4x^5
a = 4x^2, b = 1 gives 4x^2 + 1 > = 4 x

adding we get

8x^10 + 2x^2+ 8x + 2 >=4x^9 + 8x^5 + 2

dividing by 2 we get 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x

2011/001) a problem in inequality

Let a,b,c,d be positive real numbers such that abcd = 1. Show that,?
(1 + a)(1 + b)(1 + c)(1 + d) >= 16

proof:

as a is real

(1+a) = (1-sqrt(a))^2 + 2 sqrt(a) or

1+ a >= 2 sqrt(a) ( we can also show it by AM GM inequality)

similarly

(1+b) > = 2 sqrt(b)

(1+c) > = 2 sqrt(c)

(1+d) > = 2 sqrt(d)

by multiplying

(1+a)(1+b)(1+c)(1+d)>= 16 sqrt(abcd) or > 16 as abcd = 1

2010/063) prove the inequality a+b/c+b+c/a+c+a/b>6 for a,b,c > 0 .?

this can be proved in at least 2 ways


method 1
we need to prove

(a+b)/c+(b+c)/a+(c+a)/b>6

or (a+b+c)/c + (a+b+c)/a + (a+b+ c)/b > 6

or (a+b+c) ( 1/a + 1/b+ 1/c) > 9

we know by AM GM inequality (a+b + c)/ 3 >= (abc)^(1/3)

again (1/a+ 1/b+ 1/c)/3 > = (1/abc)^(1/3)

multiplying we get (a+b + c)/ 3 * (1/a+ 1/b+ 1/c)/3 > = 1

or (a+b + c) * (1/a+ 1/b+ 1/c) >= 9

it is equal when a = b = c

proved


method 2
x+ 1/x >= 2 as (x+1/x) = (sqrt(x) -1/sqrt(x))^2 + 2

puttiing x = a/b . b/c and c/a in 3 times we get 3 eqautions

a/b + b/a >= 2
a/c+ c/a > = 2
b/c + c/b > = 2
adding above 3 and rearranging the terms we get the result.

2008/008) Tough inequality

Prove that for all x>0, y>0, and all real a it holds true that

(x^((sin(a))^2))*(y^((cos(a))^2))
let us assume y > x

x^t y^(1-t) < x+ y putting (sin a)^2 = t we get (cos a)^2 = 1-t

devide by x on both sides

x^(t-1)y^(1-t) < (1+y/x)

or (y/x)^(1-t) < 1+ y/x

or m ^k < 1+m where k < 1 and m > 1
we know m ^k < m where k < 1

so m^ k < 1+m

if x > y then role of sin a and cos a are reversed and you get the same result