MHB What is the Maximum Value of 1/𝑥 + 1/𝑦 with x+y=5 and Positive Integers?

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The maximum value of 1/x + 1/y, given that x + y = 5 and both x and y are positive integers, can be determined by evaluating the possible pairs of integers that sum to 5. The pairs (1,4), (2,3), (3,2), and (4,1) yield different values for the expression. After calculating, the pair (2,3) or (3,2) provides the highest value of 1/x + 1/y, which is 5/6. This approach involves systematically testing each integer combination to find the optimal solution. The maximum value of the expression is thus 5/6.
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If 𝑥+𝑦=5 and 𝑥 and 𝑦 are positive integers, then the largest possible value of 1/𝑥 + 1/y is
 
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prasadini said:
If 𝑥+𝑦=5 and 𝑥 and 𝑦 are positive integers, then the largest possible value of 1/𝑥 + 1/y is
There are not many ways of expressing 5 as the sum of two positive integers. I suggest that you look at each of them in turn, and find which one gives the best result for 1/𝑥 + 1/y.
 
Basically, you have to do two additions of fractions.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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