Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question: Prove that a/(b+c)+b/(c+a)+c/(a+b)>=3/2, for each a,b,c>0.I'm not sure how to use Lagrange multipliers for it... any ideas?

Brugolino7t

Brugolino7t

Open question

2022-08-20

Proving inequality using Lagrange multipliers.
I started learing about Lagrange Multipliers and I got the following question:
Prove that a b + c + b c + a + c a + b 3 2 , for each a , b , c > 0
I'm not sure how to use Lagrange multipliers for it... any ideas?

Answer & Explanation

Branson Grimes

Branson Grimes

Beginner2022-08-21Added 9 answers

Since our inequality is homogeneous, we can assume a + b + c = 3
Also let f ( a , b , c , λ ) = c y c a 3 a + λ ( a + b + c 3 ) 3 2
f a = 3 ( 3 a ) 2 + λ
Let in ( a , b , c ) our f gets a minimal value ( f is a continuous function and this thing happens on the compact).
Thus, in this point f a = f b = f c = 0, which gives
3 ( 3 a ) 2 = 3 ( 3 b ) 2
and since a + b < 3, we obtain: a = b, which says that c = 3 2 a and it remains to prove that
2 a 3 a + 3 2 a 2 a 3 2 ,
where 0 < a < 3 2 , which gives ( a 1 ) 2 0
Also for a 0 + ... our inequality is obviously true.
Done!
Memphis Khan

Memphis Khan

Beginner2022-08-22Added 1 answers

Let f ( x , y , z ) = x y + z + y x + z + z x + y
f x ^ = + 1 ( y + z ) y ( x + z ) 2 z ( x + y ) 2 f y ^ = x ( y + z ) 2 + y ( x + z ) z ( x + y ) 2 f z ^ = x ( y + z ) 2 y ( x + z ) 2 + z ( x + y )
You should be able to demonstrate that
f = 0 x = y = z
and since f ( x , x , x ) = 3 2 you are pretty well done.

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