# Inequality with Two Minima

### Problem

### Remark

$x\lt \min\{y,z\}\;$ exactly means that $x\;$ is less than both $y\;$ and $z,\;$ i.e., that $x\lt y\;$ and $x\lt z.\;$ It follows that the problem is equivalent to proving two independent inequalities:

(*)

$\displaystyle\sqrt{a-c}+\sqrt{b-c}\le\sqrt{\frac{ab}{c}}$

and

(**)

$\sqrt{a-c}+\sqrt{b-c}\le\sqrt{2(a+b-c)}.$

### Proof 1

Inequality (*) is equivalent to

$\displaystyle\sqrt{\frac{c}{b}\cdot\frac{a-c}{a}}+\sqrt{\frac{c}{a}\cdot\frac{b-c}{b}}\le 1.$

Indeed, the AM-GM inequality yields

$\displaystyle\begin{align} \sqrt{\frac{c}{b}\cdot\frac{a-c}{a}}+\sqrt{\frac{c}{a}\cdot\frac{b-c}{b}} &\le \frac{1}{2}\left(\frac{c}{b}+\frac{a-c}{a}\right)+\frac{1}{2}\left(\frac{c}{a}+\frac{b-c}{b}\right)\\ &=\frac{1}{2}\left[\left(\frac{a-c}{a}+\frac{c}{a}\right)+\left(\frac{b-c}{b}+\frac{c}{b}\right)\right]\\ &=\frac{1}{2}(1+1)\\ &=1. \end{align}$

Equality occurs if $\displaystyle\frac{c}{b}=\frac{a-c}{a}\;$ and $\displaystyle\frac{c}{a}=\frac{b-c}{b},\;$ both taking place if $\displaystyle c=\frac{ab}{a+b}.$

Inequality (**) is shown directly with the AM-QM inequality:

$\sqrt{a-c}+\sqrt{b-c}\le\sqrt{2[(a-c)+(b-c)]},$

with equality if $a-c=b-c,\;$ if $a=b.$

### Proof 2

For (*), observe that, by the Cauchy-Schwarz inequality

$\displaystyle\begin{align} \sqrt{\frac{ab}{c}} &= \frac{\sqrt{ab}}{\sqrt{c}}\\ &= \frac{ \sqrt{[(a-c)+c][c+(b-c)]}}{\sqrt{c}}\\ &\ge\frac{\sqrt{\left[\sqrt{(a-c)c}+\sqrt{c(b-c)}\right]^2}}{\sqrt{c}}\\ &=\frac{\sqrt{(a-c)c}+\sqrt{c(b-c)}}{\sqrt{c}}\\ &=\sqrt{a-c}+\sqrt{b-c}. \end{align}$

Inequality (**) is shown via Jensen's inequality for $f(x)=\sqrt{x}:$

$\displaystyle\frac{\sqrt{a-c}+\sqrt{b-c}}{2}\le\sqrt{\frac{(a-c)+(b-c)}{2}}.$

### Proof 3

Note that

$\displaystyle\sqrt{x}+\sqrt{y}\le\sqrt{x+y+\frac{xy}{z}+z}=\sqrt{\frac{(x+z)(y+z)}{z}}.$

With $x=a-c,\;y=b-c,\;$ and $z=c,\;$ we get (*).

For (**), by the Cauchy-Schwarz inequality,

$2(a+b-c)=[1+1][(a-c)+(b-c)]\ge \left(\sqrt{a-c}+\sqrt{b-c}\right)^2.$

### Proof 4

WLOG, $c\le a\le b\;$ such that $a=c+x,\;$ $b=c+x+y,\;$ $x+y\gt 0.$

For (*), we wish to prove $\sqrt{cx}+\sqrt{c(x+y)}\le\sqrt{(x+c)(c+x+y)},\;$ or equivalently (by squaring),

$cx+cx+cy+2c\sqrt{x(x+y)}\le cx+x^2+xy+c^2+cx+yx$

which simplifies to $c^2-2c{x(x+y)}+x(x+y)\ge 0,\;$ i.e., $(c-\sqrt{x(x+y)})^2\ge 0\;$ which is true. Equality is achieved when $c^2=x(x+y),\;$ i.e., when $c^2=(a-c)(b-c),\;$ or $\displaystyle c=\frac{ab}{a+b}.$

Observe now that (**) is equivalent to $\sqrt{x}+\sqrt{x+y}\le\sqrt{2(2x+y)}\;$ which, by squaring, reduces to $x+x+y+2\sqrt{x(x+y)}\le 4x+2y,\;$ i.e., $x+(x+y)-2\sqrt{x(x+y)}\ge 0,\;$ or, finally, $\left(\sqrt{x+y}-\sqrt{x}\right)^2\ge 0\;$ which is true. Equality when $y=0,\;$ i.e. when $a=b.$

### Proof 5

With $\displaystyle x=\frac{a}{c}\;$ and $\displaystyle y=\frac{b}{c},\;$ (*) becomes

$\sqrt{x-1}+\sqrt{y-1}\le\sqrt{xy}.$

Squaring gives $x-1+y-1+2\sqrt{(x-1)(y-1)}\le xy,\;$ and, subsequently, $2\sqrt{(x-1)(y-1)}\le (x-1)(y-1)+1,\;$ which is of the form $2X\le X^2+1,\;$ which is true.

(**) is shown as in Proof 1.

### Acknowledgment

Dorin Marghidanu has posted the problem and a solution (Proof 1) at the CutTheKnotMath facebook page, with additional solutions by Marian Dinca (Proof 2), by (Proof 3), by Imad Zak (Proof 4).

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