Let a 1 , a 2 , a 3 … … a n be an A P. Prove that : 1 a 1 a n + 1 a 2 a n - 1 + 1 a 3 a n - 2 + … … … + 1 a n a 1 = 2 a 1 + a n 1 a 1 + 1 a 2 + 1 a 3 + … … … + 1 a n

Given, a 1 , a 2 , . . . . a n are in AP In AP , T n = a + ( n - 1 ) d ⇒ T r = a + ( r - 1 ) d & T n - r + 1 = a + ( n - r ) d ⇒ T r + T n - r + 1 = 2 a + ( n - 1 ) d which is independent of r ⇒ T 1 + T n = T 2 + T n - 1 = . . . = T n + T 1 ⇒ a 1 + a n = a 2 + a n - 1 = a 3 + a n - 2 . . . = a n - 1 + a 2 = a n + a 1 → 1 Let us consider LHS ; 1 a 1 a n + 1 a 2 a n - 1 + 1 a 3 a n - 2 + … … … + 1 a n a 1 multiply ( a 1 + a n ) in numerator and denominator for every term of the above equation = 1 a 1 + a n a 1 + a n a 1 a n + a 1 + a n a 2 a n - 1 + a 1 + a n a 3 a n - 2 + … . + a 1 + a n a n a 1 After using 1 ; = 1 a 1 + a n a 1 + a n a 1 a n + a 2 + a n - 1 a 2 a n - 1 + a 3 + a n - 2 a 3 a n - 2 + … . + a 1 + a n a n a 1 = 1 a 1 + a n 1 a n + 1 a 1 + 1 a 2 + 1 a n - 1 … … + 1 a 1 + 1 a n = 2 a 1 + a n 1 a 1 + 1 a 2 + 1 a 3 … … + 1 a n

Prove that : ( a b + x y ) ( a x + b y ) ≥ 4 a b x y where a , b , x , y ∈ R +

Use AM ≥ GM inequality Consider two terms a b & x y ; Apply AM ≥ GM a b + x y 2 ≥ ( a b x y ) 1 / 2 . . . i Consider two terms a x & b y ; Apply AM ≥ GM a x + b y 2 ≥ ( a b x y ) 1 / 2 . . . i i Multiply equations i × ( i i ) ⇒ ( a b + x y ) 2 ( a x + b y ) 2 ≥ a b x y 1 2 a b x y 1 2 ⇒ ( a b + x y ) ( a x + b y ) ≥ 4 a b x y

If a , b , c ∈ R & a + b + c = 1 ; then show that ( 1 - a ) ( 1 - b ) ( 1 - c ) ≥ 8 a b c

Given a + b + c = 1 → ( 1 ) Let us consider ( a & b ) and apply the relation AM ≥ GM ; a + b 2 ≥ ( a b ) 1 / 2 Replace a + b with 1 - c from 1 ( 1 - c ) ≥ 2 ( a b ) 1 / 2 . . . i Similarly ( 1 - a ) ≥ 2 ( b c ) 1 / 2 . . . i i ( 1 - b ) ≥ 2 ( a c ) 1 / 2 . . . i i i by Multiplying ( i ) ×( i i )×( i i i ) , we get ( 1 - a ) ( 1 - b ) ( 1 - c ) ≥ 8 a b c

For positive numbers a , b , c show that b c a + a c b + a b c ≥ a + b + c

Consider two terms b c a & a c b Apply the relation AM ≥ GM to above; b c a + a c b 2 ≥ a b c 2 a b 1 / 2 ⇒ b c a + a c b ≥ 2 c → i Similarly, for the terms a c b & a b c ; a c b + a b c ≥ 2 a → i i AM ≥ GM Similarly for the terms a b c & b c a a b c + b c a ≥ 2 b → i i i Now add ( i ) , ( i i ) & ( i i i ) , 2 a b c + b c a + c a b ≥ 2 ( a + b + c ) ⇒ a b c + b c a + c a b ≥ ( a + b + c )