The real numbers a,b,c,da, b, c, d are such that abcd>0a \geq b \geq c \geq d > 0 and a+b+c+d=1a + b + c + d = 1. Prove that (a+2b+3c+4d)aabbccdd<1.(a + 2b + 3c + 4d) a^a b^b c^c d^d < 1.