Solving the 1/a + 1/b + 1/c = 3 Equation

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The equation 1/a + 1/b + 1/c = 3 can be solved with the solution a = b = c = 1, which satisfies the condition a, b, c > 0. For values greater than 1/3, it is established that a, b, c must be greater than 1/3 to maintain the equation's balance. If a, b, and c are whole numbers, they must all equal 1, as any other values would result in fractions. The discussion also hints at a relationship between the two equations involving roots, suggesting further exploration is needed. The key takeaway is that the simplest solution under the given constraints is a = b = c = 1.
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a,b,c >0

1/a + 1/b + 1/c = 3

How can solve that
⁴√(a³) + ⁴√(b³) + ⁴√(c³) ≥ ³√(a²) + ³√(b²) + ³√(c²)

thank you very much!
 
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Well, I can see that one solution is a=b=c=1 (you probably already know that, though :wink:).

Past that, I don't know. One number must be greater than one, one less, and one relates to the other two by the second formula.
 
c6_viyen_1995 said:
a,b,c >0

1/a + 1/b + 1/c = 3

How can solve that
⁴√(a³) + ⁴√(b³) + ⁴√(c³) ≥ ³√(a²) + ³√(b²) + ³√(c²)

thank you very much!

If you start with 1/a=3, you find a = 1/3.
With 1/a + 1/b = 3, you can see that a > 1/3, because b > 0.
Due to the symmetry of a and b, that means also that b > 1/3.

With 1/a + 1/b + 1/c = 3, it follows that a,b,c > 1/3.

I suspect that is the intended answer.

Otherwise, if a,b,c are supposed to be whole numbers, it follows that a,b,c ≥ 1.
It follows that then a = b = c = 1, because otherwise we would have fractions.


The second equation can be "solved" equivalently. Do you see how?
 

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