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For the question below I have split the inequalities into two and currently have

LS $$\frac{1}{4} k^4 + k^3 < \frac{1}{4} (k+1)^4$$

RS $$\frac{1}{4} (k+1)^4 < \frac{1}{4} k^4 + (k+1)^3$$

I am unsure of what I am solving for after this. When one side equals the other it makes sense that I have to make the two sides the same. With inequalities however how can I show one is less then the other? What am I trying to show?

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user490305
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  • Multiply out the parantheses and since some terms cancel both inequalities can relatively easy be verified. – Peter Oct 17 '17 at 17:07

2 Answers2

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HINT: use that $$\sum_{i=1}^{n-1}i^3=\frac{1}{4} (n-1)^2 n^2$$

Dr. Sonnhard Graubner
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We should prove that

$${ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }<\frac { { \left( n+1 \right) }^{ 4 } }{ 4 } <{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+\left( n+1 \right) ^{ 3 }$$

$$L.H.\quad \underset { <\frac { { n }^{ 4 } }{ 4 } }{ \underbrace { { 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ \left( n-1 \right) }^{ 3 } } +{ n }^{ 3 } } <\frac { { n }^{ 4 } }{ 4 } +{ n }^{ 3 }=\frac { { n }^{ 4 }+4{ n }^{ 3 } }{ 4 } <\frac { { n }^{ 4 }+4{ n }^{ 3 }+6{ n }^{ 2 }+4n+1 }{ 4 } =\frac { { \left( n+1 \right) }^{ 4 } }{ 4 } \\ R.H.{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }+\left( n+1 \right) ^{ 3 }>\frac { { n }^{ 4 } }{ 4 } +\left( n+1 \right) ^{ 3 }=\frac { { n }^{ 4 }+4\left( n+1 \right) ^{ 3 } }{ 4 } =\frac { { n }^{ 4 }+4{ n }^{ 3 }+12{ n }^{ 2 }+12n+4 }{ 4 } >\frac { { n }^{ 3 }+4{ n }^{ 3 }+6{ n }^{ 2 }+4n+1 }{ 4 } =\frac { { \left( n+1 \right) }^{ 4 } }{ 4 } \\ $$

haqnatural
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