A Curious Inequality with... Radicals // ... रेडिकलहरूसँगको एउटा जिज्ञासु असमानता

 The page presents the problems from the fifth onwards proposed for the 7th grade (the first four are on the previous page)

As it says on the second cover: "In grades VII-XII, the first 8 problems are intended for deepening the subject and preparing for national exams, and the last 4 are addressed to those who wish to additionally prepare for competitions". 

So Problem S:E25.203 is considered easy. But still... ??

         In translation : 

    S:E25.203.  The positive real numbers  $a$  and  $b$  verify the relation  $a+b+ab=3.$

                a) Verify that, if  $a=2$  then  $a+b>2>\sqrt{a}+\sqrt{b}.$

                b)  Prove that  $a+b\geqslant \sqrt{a}+\sqrt{b}$,  for any  $a$  and  $b.$

ANSWER CiP

$\textbf{b)}\;\;\;  a+b\geqslant 2\geqslant \sqrt{a}+\sqrt{b} \tag{A}$.

 The  $"="$ sign occurs if and only if  $a=b=1$

                         Solution CiP

               a) From the given relation we obtain

$b=\frac{3-a}{1+a} \tag{1}$

and it is observed that the given numbers are within the range of values  $0<a,\;b<3$.

          $a=2\underset{(1)}{\Rightarrow}b=\frac{1}{3}$  and obviously  $a+b>2$.

     After that, we have  $\sqrt{a}+\sqrt{b}=\sqrt{2}+\sqrt{\frac{1}{3}}=\frac{3\sqrt{2}+\sqrt{3}}{3}.$  But

$(\sqrt{a}+\sqrt{b})^2=\frac{(3\sqrt{2}+\sqrt{3})^2}{9}=\frac{18+3+6\sqrt{6}}{9}=\frac{7+2\sqrt{6}}{3}<4$, because of  

$\sqrt{24}<\sqrt{25}\Rightarrow 2\sqrt{6}<5\Rightarrow 7+2\sqrt{6}<7+5 \Rightarrow \frac{7+2\sqrt{6}}{3}<4$.

Interestingly, we don't get a more precise result if we start with the "stronger" inequality  $\sqrt{288}<\sqrt{289}$. So

$2<\frac{289}{144}\Rightarrow \sqrt{2}<\sqrt{\frac{289}{144}}=\frac{17}{12}\Rightarrow 4\sqrt{2}<\frac{17}{3}\Rightarrow 6-4\sqrt{2}>6-\frac{17}{3}=\frac{1}{3}\Rightarrow \sqrt{4-4\sqrt{2}+(\sqrt{2})^2}>\sqrt{\frac{1}{3}}\Rightarrow$

$\Rightarrow \sqrt{(2-\sqrt{2})^2}>\sqrt{\frac{1}{3}}\Rightarrow 2-\sqrt{2}>\sqrt{\frac{1}{3}} \Rightarrow 2>\sqrt{2}+\sqrt{\frac{1}{3}}$.

               b)  Only after numerous numerical simulations was I led to the solution of this case, by introducing the intermediate term  $2$  between the two inequalities.

        In the first part we use

  $x+\frac{4}{x}\geqslant 4,\;for\;x>0 \tag{2}$

 ($\Leftrightarrow (\sqrt{x})^2-2\cdot \sqrt{x} \cdot \frac{2}{\sqrt{x}}+\left (\frac{2}{\sqrt{x}}\right )^2\geqslant 0\Leftrightarrow \left (\sqrt{x}-\frac{2}{\sqrt{x}}\right )^2\geqslant 0;\;"="\;iff\;\;x=2$)

We have  

$a+b\underset{(1)}{=}a+\frac{3-a}{1+a}=(a-1)+\left ( \frac{3-a}{1+a}+1 \right )=(a-1)+\frac{4}{1+a}=$

$=\left ((1+a)+\frac{4}{1+a}\right )-2\;\overset{(2)}{\underset{with\;x=1+a}{\geqslant}}\;4-2=2$

thus left side of (A). Equal sign occurs here iff  $1+a=2$  so  $a=1$  and hence  $b\overset{(1)}{=}\frac{3-1}{1+1}=1$.

For right side of (A),  $(\sqrt{ab}-1)^2\geqslant 0\Rightarrow ab-2\sqrt{ab}+1\geqslant 0 \;\overset{ab=3-a-b}{\Rightarrow}$

$\Rightarrow 3-a-b-2\sqrt{ab}+1\geqslant 0\Rightarrow a+b+2\sqrt{ab}\leqslant 4\Rightarrow (\sqrt{a}+\sqrt{b})^2\leqslant 4$

so  $\sqrt{a}+\sqrt{b}\leqslant 2$. The  $"="$  sign occurs if and only if  $a\cdot b=1$  which together with  $a+b=3-ab=2$  give  $a=b=1$.

$\blacksquare$