Small induction problem.

Question

Show with the help of induction for all natural numbers n, where a and b are real numbers and: $a \neq b$ and $ a + b > 0$

$$(a + b)^n ≤ 2^{n−1}\cdot(a^n + b^n) $$

Answer 1

Induction is to be performed on $n$, with $a,b$ fixed.So we start with $n = 0$, which gives $1$ on the LHS and $\frac 12 \times 2 = 1$ on the RHS. So the statement is true for $n = 0$. Similarly you can check $n = 1$ as well.

For the induction step, you must assume that $(a+b)^{k} \leq 2^{k-1} \times (a^k + b^k)$, and get the same statement with $k$ replaced by $k+1$.

For this, we first ensure the left hand side is converted to the case $k+1$. For this, multiply by $a+b$(which is positive, so the inequality is preserved) to get $(a+b)^{k+1} \leq 2^{k-1}(a^k + b^k)(a+b)$. But we want the right hand side to be $2^k(a^{k+1} + b^{k+1})$ so that we can be done.

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For this, let us first expand $2^{k-1}(a^k+b^k)(a+b)$. It is equal to $2^{k-1}(a^{k+1}+b^{k+1} + a^k b + b^ka)$.

Suppose we show that (call this statement $(*)$) $a^kb + b^k a \leq a^{k+1} + b^{k+1}$, then the expansion is less than or equal to $2^{k-1} \times 2 \times (a^{k+1} + b^{k+1})$, and we would be done by combining the powers of $2$.

Try to prove $(*)$ yourself. See the solution below if you cannot.

To show $(*)$, we first write down the RHS minus the LHS : it is $a^{k+1} - a^k b + b^{k+1} - b^ka$. This gives $a^k(a-b) + b^k(b-a)$, which then simplifies to $(b^k - a^k)(b-a)$, which is positive : if $a>b$ then both terms are negative, else both terms are positive, either way the product is positive. So RHS minus LHS is positive, this is $(*)$.

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Hence we are done.