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anemone
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If $a$ and $b$ are positive real numbers, show that $a^b+b^a>1$.
anemone said:If $a$ and $b$ are positive real numbers, show that $a^b+b^a>1$.
note the values of :$a^b\, and \,\,\, b^a$ are variables ,not fixedmathbalarka said:I don't think that is a correct argument.
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.
in fact if a=0 or b=0 then :mathbalarka said: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 $b = 4$.
Albert said: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
are you sure we can not find solutionskaliprasad said:not correct because g(1) = 0 is wrong pressumption
[sp]First, I want to prove the inequality $x^x > 1-\sqrt x$, for $0<x<1.$ Taking logs, this is equivalent to $x\ln x > \ln(1-\sqrt x).$anemone said:If $a$ and $b$ are positive real numbers, show that $a^b+b^a>1$.
Nice proof!kaliprasad said: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 $a^b+b^a \gt 1$
kaliprasad said:or $\dfrac{a+b}{b} \gt (\dfrac{1}{b})^a$
or $\dfrac{b}{a+b} \gt b^a$
Just a typo – a couple of the inequality signs are the wrong way round.I like Serena said:How is that?
I like Serena said:How is that?
Opalg said:[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, \(\displaystyle g'(x) = \frac1x - \frac1{4x^{3/2}},\) 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 said: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 $a^b+b^a \gt 1$
The inequality $a^b+b^a > 1$ is an important result in mathematics and has implications in many fields such as number theory, geometry, and physics. It is also a fundamental concept in understanding the behavior of exponential and power functions.
There are multiple approaches to proving this inequality. One way is to use the properties of logarithms and the AM-GM (arithmetic mean-geometric mean) inequality. Another approach is to use the concept of convexity and the Bernoulli's inequality. Both methods involve algebraic manipulations and the use of basic inequalities.
No, this inequality only holds for positive real numbers $a,b$. For complex numbers, there is no clear definition of exponentiation and the concept of inequality is not well-defined.
Yes, when $a=b=1$, the inequality becomes $1+1=2 > 1$, which is true. However, for any other positive real numbers $a,b$, $a^b+b^a$ will always be greater than 1.
This inequality is closely related to the concept of convexity, the AM-GM inequality, and Bernoulli's inequality. It also has connections to the concept of growth rates and asymptotic behavior in calculus.