TargetIIT

Just substitute

a=1+A
b=1+B
c=1+C
d=1+D
AS they all are >1

: =  A,B,C,D >0

then we need to prove

8((1+A)(1+B)(1+C)(1+D)+1)>(A+2)(B+2)(C+2)(D+2)

=>

8(ABCD+ΣABC+ΣAB+ΣA+2)>16((A/2)(B/2)(C/2)(D/2)+ΣABC/8
                                            +(ΣAB/4)+ΣA/2+1)

simplifying ,

8ABCD+8ΣABC+8ΣAB+8ΣA+16>ABCD+2ΣABC+4ΣAB+8ΣA+16

which can be clearly seen

WLOG a[im]http://codecogs.izyba.com/gif.latex?a%20%5Cge%20b%20%5Cge%20c%20%5Cge%20d[/im]

The sequences (1,1) and (a,b) are similarly ordered.

So, from Chebyshev's Inequality, we have 2(1+ab) >=(1+a)(1+b)

Similarly, we get 2(1+cd) >= (1+c)(1+d)

Again, the sequences (1,1) and (ab,cd) are similarly sorted.

So, 2(1+abcd) >= (1+ab)(1+cd)

Hence 8(1+abcd) >= 2(1+ab) 2(1+cd) >= (1+a)(1+b)(1+c)(1+d)

I dont think so. I havent seen any real use of inequalities in JEE beyond AM-GM.

The standard ones in Integrals (Riemann sums), and m(b-a) <= I <= M(b-a) those are used at times

well we see that

a[p]2[/p]+b[p]2[/p]/a+b >(a+b)/2 as

It reduces to


a[p]2[/p]+b[p]2[/p]>2ab which is true as (a-b)[p]2[/p]>0

so applying it on the other two terms we get

>= (a+b)/2 +(b+c)/2 + (c+a)/2=a+b+c