Prove that ac 3bd2 Solution6 ac 3bd2 3ad2 3bc2 ad bc2 0

Solution

6) . (ac + 3bd)^2 <= (a^2 +3b^2) (c^2 +3d^2)

(ac)^2 + (3bd)^2 +2ac(3bd) = (ac)^2 + (3bd)^2 + 3(ad)^2 + 3(bc)^2

now

6abcd < = 3(ad)^2 + 3(bc)^2

-6abcd > = -3(ad)^2 -3(bc)^2

(ad -bc)^2 > =0

ad >bc

(a² + 3b²)(c² + 3d²) = a² (c² + 3d²) + 3b² (c² + 3d²) =
= a²c² + 3a²d² + 3b²c² +9b²d² =
= a²c² - 6abcd + 9b²d² + 3a²d² + 6abcd + 3b²c² =
= (ac +3bd)² + 3(ad - bc)²

^{now just keep} (a² + 3b²)(c² + 3d²) = (ac +3bd)² + 3(ad - bc)²

the inequality is (ac +3bd)² <= (a² + 3b²)(c² + 3d²

(ac +3bd)²    <= (ac +3bd)² + 3(ad - bc)²

^{by this we can easily say that on right hand side we have a term 3(ad -bc)^2 term extra , so this will add upto some value}

^{hence proved}