By Holder's inequality ,
a + √ a b + 3√ a b c ≤ { a + a + a } 1 / 3 { a + √ a b + b } 1 / 3 { a + b + c } 1 / 3
≤ 3 2 / 3 { a + √ a b + b } 1 / 3 { 2a + b } 1 / 3 { 3√a (a + b ) ( a + b + c )6 }
It remains to show that ,
3 5 / 2 { a + √ a b + b } 1 / 3 { 2a + b } 1 / 3 ≤ 8 + 2 √ a ba + b
Or equivalently ,
3 5 { 2 + 2 √ a ba + b } ≤ { 8 + 2 √ a ba + b } 3
Let x = 2 √ a ba + b .
Using AM - GM ,
3 5 { 2 + x } = 3 3 [ 3 . 3 . { 2 + x } ] ≤ 3 3 { 3 + 3 + 2 + x3 } 3 = { 8 + x } 3
You could look up Wikipedia for a slightly modified inequality of this one , but I forgot its name .