Dorin Marghidanu's Calculus Lemma

Proof 1

Indeed, passing to logarithms, we obtain equivalently

$(v-1)\ln u+\ln (u+v)\gt 0.$

To prove that, consider the function $f:\;(0,\infty)\rightarrow\mathbb{R},\;$ defined by $f(x)=(v-1)\ln x+\ln (x+v),\;$ for which we have $\displaystyle f'(x)=v\frac{x+v-1}{x(x+v)}.\;$ Hence, $f\;$ has a lone minimum at $x_0=1-v\in (0,1).\;$ Therefore, for any $x\in (0,\infty),\;$ $f(x)\gt f(x_0),\;$ i.e.,

$f(x)=(v-1)\ln x+\ln (x+v)\gt (v-1)\ln (1-v)\gt 0,$

which proves the lemma.

Proof 2

We make use of Bernoulli's inequality:

If $x\gt -1\;$ and $0\lt a\lt 1,\;$ then $(1+x)^a\lt 1+ax.$

Apply Bernoulli's inequality with $x=u-1\ge -1\;$ and $a=1-v\in (0,1),\;$ to obtain

\begin{align} u^{1-v} &= (1+(u-1))^{1-v}\\ &\lt 1+(1-v)(u-1)\\ &=u+v-uv\\ &\lt u+v. \end{align}

Hence, $\displaystyle u^v\gt\frac{u}{u+v}.$

Proof 3

We'll use another form of Bernoulli's inequality:

For $h\gt 0$ and $a\gt 1,$

$(1+h)^a\gt 1+ah.$

Set $\displaystyle h=\frac{v}{u}$ and $\displaystyle a=\frac{1}{v}.$ Then $\displaystyle (1+\frac{v}{u})^{\frac{1}{v}}\gt 1+\frac{v}{u}\cdot\frac{1}{v}\gt \frac{1}{u}.$ It then follows that

$\displaystyle \frac{u+v}{u}\gt\frac{1}{u^v},$

or, $\displaystyle u^v\gt \frac{u}{u+v}.$

Applications

Here's a list of simple consequences of the above lemma:

1. If $a,b\in (0,1),\;$ then

$\displaystyle a^b+b^a\gt\frac{a}{a+b}+\frac{b}{b+a}=1.$

2. If $a,b,c\in \displaystyle (0,\frac{1}{2}),\;$ then

$\displaystyle a^{b+c}+b^{c+a}+c^{a+b}\gt\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}=1.$

3. If $a,b,c\in \displaystyle (0,1),\;$ then

$\displaystyle (a+b)^c+(b+c)^a+(c+a)^b\gt\frac{a+b}{a+b+c}+\frac{b+c}{b+c+a}+\frac{c+a}{c+a+b}=2.$

4. For $x_k,$ $k=1,2,\ldots,n,$ all positive,

$\displaystyle \sum_{k=1}^n(x_1+\ldots+x_{k-1}+x_{k+1}+\ldots+x_n)^{x_k}\gt n-1.$

(This is problem M1219 from Kvant referred to below.)

5. If $x\in\displaystyle\left(0,\frac{\pi}{2}\right),\;$ then

$\displaystyle \sin x^{\cos x}+\cos x^{\sin x}\gt\frac{\sin x}{\sin x+\cos x}+\frac{\cos x}{\cos x+\sin x}=1.$

6. If $a,b,c,d\in (0,1),\;$ then

$\displaystyle a^b\cdot (a+b)^c\cdot (a+b+c)^d\gt\frac{a}{a+b+c+d}.$

For a proof, see a separate page.

7. If $a,b,c\in (0,1),\;$ then

$2^a(b+c)^{1-a}+2^b(c+a)^{1-b}+2^c(a+b)^{1-c}\lt 4(a+b+c).$

For a proof, see a separate page.

8. If $a,b,c,d\in (0,1),\;$ then

$\displaystyle \left(\frac{a+b}{2}\right)^{\left(\frac{c+d}{2}\right)}+\left(\frac{b+c}{2}\right)^{\left(\frac{d+a}{2}\right)}+\left(\frac{c+d}{2}\right)^{\left(\frac{a+b}{2}\right)}+\left(\frac{d+a}{2}\right)^{\left(\frac{b+c}{2}\right)} \gt 2.$

Acknowledgment

The lemma, with the first two proofs and a list of applications, has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Proof 3 appeared in the Russian Kvant magazine (n 9, 1990) in the solution of problem M1219.