Inequalities
If a , b , c ∈ R + . Then the minimum value of c a + b + a b + c + b c + a = ?
The answer is 6.
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4 solutions
Moderator note:
The minimum is the greatest lower bound .
You still need to show that equality can occur. Otherwise, all that you have is a lower bound, which may not be the minimum. For example, it is obvious that each term is ≥ 0 and so the sum is ≥ 0 , but we cannot conclude that the minimum is 0. All that we have is that 0 is a lower bound.
Again a , b , c need not be distinct. In fact equality occurs ⟺ a = b = c .
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Disagree with "simple standard solution" for exactly the reason that you stated.
Yup, thanks for that. Edited.
Here is my solution:
We shall normalize the inequality by substituting a + b + c = 1 . So, the expression reduces to a 1 − a + b 1 − b + c 1 − c now, let f ( x ) = x 1 − x . Since this function is convex, applying Jensen's Inequality, we have f ( 3 a + b + c ) ≤ 3 f ( a ) + f ( b ) + f ( c ) this implies that 2 × 3 ≤ a 1 − a + b 1 − b + c 1 − c and hence the minimum value of this expression is 2 × 3 = 6
I am sorry to post the solution in the comment box. I posted it here because there was an error when I typed the solution in the answer box. Sorry Brilliant if I have done anything wrong.
b a + a b + c a + a c + c b + b c ≥ 2 + 2 + 2 = 6
Put a,b,c=1then very easily we calculate the ans..
Applying AM GM inequality on < c a , c b , a b , a c , b c , b a > we get;
6 1 ( c a + c b + a b + a c + b c + b a ) ≥ 6 c a × c b × a b × a c × b c × b a c a + b + a b + c + b c + a ≥ 6
Let's assume a = b = c = 1 ∴ c a + b + a b + c + b c + a = 1 1 + 1 + 1 1 + 1 + 1 1 + 1 = 2 + + 2 + 2 = 6