Proving $a^b+b^a > 1$ for Positive Real $a,b$

[anemone]

If $a$ and $b$ are positive real numbers, show that $a^b+b^a>1$.

 

[Albert1]

If $a$ and $b$ are positive real numbers, show that $a^b+b^a>1$.

Spoiler

$a^b>0,b^a>0 $ are the roots of the following equation :
$(x-a^b)(x-b^a)=0$
or $x^2-x(a^b+b^a)+a^bb^a=0$
let x=1, we have:
$1-(a^b+b^a)+a^bb^a=0$
$\therefore 1+a^bb^a=(a^b+b^a), \,\, and \,\, the\,\, proof \,\,is \,\, done$

 

[mathbalarka]

I don't think that is a correct argument.

Spoiler

For example, $2$ and $3$ are roots of the equation $(x - 2)(x - 3) = x^2 - 5x + 6$ but if one sets $x = 1$ then $1 - 5 + 6 = 0$, a false equality.

 

[Albert1]

I don't think that is a correct argument.

Spoiler

For example, $2$ and $3$ are roots of the equation $(x - 2)(x - 3) = x^2 - 5x + 6$ but if one sets $x = 1$ then $1 - 5 + 6 = 0$, a false equality.

note the values of :$a^b\, and \,\,\, b^a$ are variables ,not fixed

 

[mathbalarka]

I though that didn't matter? Your equality was

$$a^b + b^a = 1 + a^b b^a$$

Which is clearly false if, say, $a = 3$ and $b = 4$.

 

[Albert1]

I though that didn't matter? Your equality was

$$a^b + b^a = 1 + a^b b^a$$

Which is clearly false if, say, $a = 3$ and $b = 4$.

in fact if a=0 or b=0 then :
$a^b+b^a= 1$
if we want to find $a^b+b^a>100$
a,b also can be found
now for all value of a>0 ,and b>0
we want to find the value of k with
$a^b+b^a>k$
if $a--->0^+ ,or\,\, b--->0^+$
we may cnclude:$a^b+b^a>1$

 

[Albert1]

Spoiler

let $g(t)=t^2-(a^b+b^a)t+(a^bb^a)$
here $ a>0 ,\, b>0$
$g(1)=1-(a^b+b^a)+a^bb^a$
now let $g(1)=0$
we have:$1-(a^b+b^a)+a^bb^a=0---(*)$
this time we will find $a,\,and \,\,b$ satisfying (*)
this can be considered as
a linear equation :$1-y+x=0$
here $y=a^b+b^a ,\,\, x=a^bb^a$
so $y=1+x$
$\therefore y=a^b+b^a=1+a^bb^a>1$
and the proof is finished

 

[kaliprasad]

Spoiler

let $g(t)=t^2-(a^b+b^a)t+(a^bb^a)$
here $ a>0 ,\, b>0$
$g(1)=1-(a^b+b^a)+a^bb^a$
now let $g(1)=0$
we have:$1-(a^b+b^a)+a^bb^a=0---(*)$
this time we will find $a,\,and \,\,b$ satisfying (*)
this can be considered as
a linear equation :$1-y+x=0$
here $y=a^b+b^a ,\,\, x=a^bb^a$
so $y=1+x$
$\therefore y=a^b+b^a=1+a^bb^a>1$
and the proof is finished

not correct because g(1) = 0 is wrong pressumption

 

[Albert1]

not correct because g(1) = 0 is wrong pressumption

are you sure we can not find solutions
$(a^b,b^a)$ satisfying :
$1-(a^b+b^a)+a^bb^a=0\,\, ?$

 

[Opalg]

If $a$ and $b$ are positive real numbers, show that $a^b+b^a>1$.

[sp]First, I want to prove the inequality $x^x > 1-\sqrt x$, for $0<x<1.$ Taking logs, this is equivalent to $x\ln x > \ln(1-\sqrt x).$

Let $f(x) = x\ln x - \ln(1-\sqrt x).$ Then $f(x)\to0$ as $x\searrow0$. So to prove the inequality it will be enough to show that $f'(x)>0$ for $0<x<1.$ But $f'(x) = 1+\ln x + \dfrac1{2\sqrt x(1-\sqrt x)} > 1+\ln x + \dfrac1{2\sqrt x}$, because $\frac 1{1 - \sqrt x} > 1.$ So it will be enough to show that $g(x)>0$, where $g(x) = 1+\ln x + \dfrac1{2\sqrt x}$.

Differentiating again, $$g'(x) = \frac1x - \frac1{4x^{3/2}},$$ which has a single zero when $x = \frac1{16}$. This turning point is a local minimum (as you can see by differentiating again), and $g\bigl(\frac1{16}\bigr) = 3-4\ln2 \approx 0.2274 > 0$. It follows that $g(x)>0$ throughout the interval $(0,1)$, as required. Working backwards, this means that the inequality $x^x > 1-\sqrt x\;\; (0<x<1)$ is proved.

Now we can look at the inequality $a^b + b^a > 1.$ It is obviously satisfied if either $a$ or $b$ is greater than $1$, so we may assume that they are both less than $1$. Next, notice that when $a,b \in (0,1)$, $a^b$ increases when $a$ increases, and it decreases when $b$ increases.

Suppose that $b < \frac12.$ Then $b^a > a^a > 1-a^{1/ 2}$ (by the above inequality). But $a^{1/ 2} < a^b$, and so $1-a^{1/ 2} > 1-a^b.$ Thus $b^a > 1-a^b$, so that the inequality $a^b+b^a >1$ is proved.

The same argument shows that the inequality holds when $a < \frac12.$ So we only have to look at the case when both $a$ and $b$ lie between $\frac12$ and $1$. But then $a^b+b^a > a+b >1.$ That completes the proof.[/sp]

 

[kaliprasad]

Spoiler

if a or b are above 1 we are done,.so let us assume both < 1
by Bernoulli Inequality -- from Wolfram MathWorld we have$(1+x)^n \ge 1+nx$so $(1+\dfrac{a}{b})^{\dfrac{1}{a}} \ge 1+\dfrac{1}{b}$so $(1+\dfrac{a}{b})^{\dfrac{1}{a}} \gt \dfrac{1}{b}$or $1+\dfrac{a}{b} \gt (\dfrac{1}{b})^a$or $\dfrac{a+b}{b} \gt (\dfrac{1}{b})^a$or $\dfrac{b}{a+b} \lt b^a$similarly

$\dfrac{a}{a+b} \lt a^b$
Adding, we get $1 \lt a^b + b^a$

or $a^b+b^a \gt 1$

 

[Opalg]

Spoiler

if a or b are above 1 we are done,.so let us assume both < 1
by Bernoulli Inequality -- from Wolfram MathWorld we have$(1+x)^n \ge 1+nx$so $(1+\dfrac{a}{b})^{\dfrac{1}{a}} \ge 1+\dfrac{1}{b}$so $(1+\dfrac{a}{b})^{\dfrac{1}{a}} \gt \dfrac{1}{b}$or $1+\dfrac{a}{b} \gt (\dfrac{1}{b})^a$or $\dfrac{a+b}{b} \gt (\dfrac{1}{b})^a$or $\dfrac{b}{a+b} \gt b^a$similarly

$\dfrac{a}{a+b} \gt a^b$
Adding, we get $1 \lt a^b + b^a$

or $a^b+b^a \gt 1$

Nice proof!

 

[I like Serena]

or $\dfrac{a+b}{b} \gt (\dfrac{1}{b})^a$

or $\dfrac{b}{a+b} \gt b^a$

How is that?

 

[Opalg]

How is that?

Just a typo – a couple of the inequality signs are the wrong way round.

 

[kaliprasad]

How is that?

I am sorry . There were typo Thanks for pointing. I have corrected the same.

 

[anemone]

[sp]First, I want to prove the inequality $x^x > 1-\sqrt x$, for $0<x<1.$ Taking logs, this is equivalent to $x\ln x > \ln(1-\sqrt x).$

Let $f(x) = x\ln x - \ln(1-\sqrt x).$ Then $f(x)\to0$ as $x\searrow0$. So to prove the inequality it will be enough to show that $f'(x)>0$ for $0<x<1.$ But $f'(x) = 1+\ln x + \dfrac1{2\sqrt x(1-\sqrt x)} > 1+\ln x + \dfrac1{2\sqrt x}$, because $\frac 1{1 - \sqrt x} > 1.$ So it will be enough to show that $g(x)>0$, where $g(x) = 1+\ln x + \dfrac1{2\sqrt x}$.

Differentiating again, $$g'(x) = \frac1x - \frac1{4x^{3/2}},$$ which has a single zero when $x = \frac1{16}$. This turning point is a local minimum (as you can see by differentiating again), and $g\bigl(\frac1{16}\bigr) = 3-4\ln2 \approx 0.2274 > 0$. It follows that $g(x)>0$ throughout the interval $(0,1)$, as required. Working backwards, this means that the inequality $x^x > 1-\sqrt x\;\; (0<x<1)$ is proved.

Now we can look at the inequality $a^b + b^a > 1.$ It is obviously satisfied if either $a$ or $b$ is greater than $1$, so we may assume that they are both less than $1$. Next, notice that when $a,b \in (0,1)$, $a^b$ increases when $a$ increases, and it decreases when $b$ increases.

Suppose that $b < \frac12.$ Then $b^a > a^a > 1-a^{1/ 2}$ (by the above inequality). But $a^{1/ 2} < a^b$, and so $1-a^{1/ 2} > 1-a^b.$ Thus $b^a > 1-a^b$, so that the inequality $a^b+b^a >1$ is proved.

The same argument shows that the inequality holds when $a < \frac12.$ So we only have to look at the case when both $a$ and $b$ lie between $\frac12$ and $1$. But then $a^b+b^a > a+b >1.$ That completes the proof.[/sp]

Thanks for participating, Opalg! Thanks for giving us such a thorough proof and I appreciate that!:)

Spoiler

if a or b are above 1 we are done,.so let us assume both < 1
by Bernoulli Inequality -- from Wolfram MathWorld we have$(1+x)^n \ge 1+nx$so $(1+\dfrac{a}{b})^{\dfrac{1}{a}} \ge 1+\dfrac{1}{b}$so $(1+\dfrac{a}{b})^{\dfrac{1}{a}} \gt \dfrac{1}{b}$or $1+\dfrac{a}{b} \gt (\dfrac{1}{b})^a$or $\dfrac{a+b}{b} \gt (\dfrac{1}{b})^a$or $\dfrac{b}{a+b} \lt b^a$similarly

$\dfrac{a}{a+b} \lt a^b$
Adding, we get $1 \lt a^b + b^a$

or $a^b+b^a \gt 1$

Hey kaliprasad,

This is one very elegant proof, well done and thanks for participating!