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

[c6_viyen_1995]

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!

 

[c6_viyen_1995]

Help me!

 

[Sciurus]

Well, I can see that one solution is a=b=c=1 (you probably already know that, though ).

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.

 

[I like Serena]

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?