For all real numbers $a$ and $b$. Prove that if $a>0$ and $b>0$, then $$ \frac{2}{a}+\frac{2}{b} \neq \frac{4}{a+b} $$

I am very confused please help , thanks.

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    $\begingroup$ what have You tried so far and which part confuses You? $\endgroup$ – Rezha Adrian Tanuharja Apr 7 at 1:01
  • $\begingroup$ Please typeset your question (in MathJax) and not post pictures, which cannot be searched. $\endgroup$ – David G. Stork Apr 7 at 1:05

Consider $$\frac2a+\frac2b -\frac4{a+b} = \frac{2(a+b)}{ab} - \frac{4}{a+b} = \frac{2(a^2+b^2+2ab)-4ab}{ab(a+b)}=\frac{2a^2+2b^2}{ab(a+b)}\ne0$$

This is because, $a,b>0$.


$$\frac2a+\frac2b \ne\frac4{a+b}$$

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  • $\begingroup$ why subtract the question $\endgroup$ – user13240088 Apr 7 at 1:08
  • $\begingroup$ To show that it is non-zero, i.e., if $x-y\ne 0$, then $x\ne y$ $\endgroup$ – Ak19 Apr 7 at 1:10
  • $\begingroup$ can be proof by contraposition? $\endgroup$ – user13240088 Apr 7 at 1:12

Suppose for purposes of contradiction that $a>0,b>0$ and that $\frac{2}{a}+\frac{2}{b}=\frac{4}{a+b}$

Multiplying both sides by $\frac{1}{2}ab(a+b)$ which we note is nonzero brings us to


Expanding and simplifying:


But since $a\neq 0$ and $b\neq 0$ the LHS is strictly positive, a contradiction.

In fact, by inspecting the proof above, we find that so long as $a\neq 0$, $b\neq 0$ and $a\neq -b$ we will never have $\frac{2}{a}+\frac{2}{b}=\frac{4}{a+b}$ which all of those conditions are already implied otherwise the fractions in the expression are undefined.

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  • $\begingroup$ can i proof by contraposition ?? $\endgroup$ – user13240088 Apr 7 at 1:24
  • $\begingroup$ @user13240088 there is not much difference between proof by contradiction and proof by contraposition. In this specific case though, I recommend against it as assuming $\frac{2}{a}+\frac{2}{b}=\frac{4}{a+b}$ will always lead to a contradiction as the conclusion you would have reached is that $a=0$, $b=0$ or that $a+b=0$ but having reached any of those conclusions means that the original expression contained undefined fractions due to division by zero errors. $\endgroup$ – JMoravitz Apr 7 at 1:28

Consider what it would mean if $\frac 2a + \frac 2b = \frac 4{a+b}$.

If we put it over a common denominator we would have:

$\frac 2a\frac {b(a+b)}{b(a+b)} + \frac 2b\frac {a(a+b)}{a(a+b)} = \frac 4{a+b}\frac {ab}{ab}$ or

$\frac {2b(a+b) + 2a(a+b)}{ab(a+b)}= \frac {4ab}{ab(a+b)}$

which would mean

$2b(a+b)+2a(a+b) = 4ab$

And if we expand that out that would mean

$2ab + 2b^2 + 2a^2 + 2ab = 4ab$ or

$4ab + 2b^2 + 2a^2 = 4ab$

and if we subtract $4ab$ from each side that would mean

$2b^2 + 2a^2 = 0$.

That would mean $b^2 = -a^2$ but as $b^2 \ge 0$ and $a^2 = 0$ that would mean $b^2 = -a^2 = 0$ and so $a=b = 0$.

But that's not possible as $\frac 2{0}$ and $\frac 4{0+0}$ is not defined.

====== Alternatively====

In general $\frac mn + \frac jk \ne \frac {m+j}{n+k}$.

For one thing $\frac {m+ j}{n+k} = \frac m{n+k} + \frac j{n+k}$ and if $m,j,n,k$ are all positive then $n< n+k, k<n+k$ so $\frac 1n > \frac 1{n+k}, \frac 1k > \frac 1{n+k}$ and $\frac mn + \frac jk > \frac m{n+k} +\frac j{n+k}$.

Of course if $m,n,j,k$ are different signs then such conclusions about $\pm\frac {|m|}{|n|\pm |k|} \pm \frac{|j|}{|n|\pm |k|}$ can not be made.

But for just $a,b$ we have the cases:

1) $a>0; b>0$ and then $\frac 2{a+b} + \frac 2{a+b} < \frac 2a + \frac 2b$.

2) $a > 0 > b > -a$. Then $\frac 2{a+b} + \frac 2{a+b}>\frac 2a + \frac 2{|b|} >\frac 2a -\frac 2{|b|} = \frac 2a + \frac 2b$.

3) $a>0>b=-a$ or $b>0>a=-b$. Then $a+b=0$ and $\frac 4{a+b}$ is not defined.

4) $a < 0; b< 0$ and that's the same as 1) just with negative values.

5) $a > 0 > -a > b$ the like 2) but but negative values.

6) $b > 0 > a> -b$ or $b>0>-b>a$ are like 2) or 4) but with the labels "$a$" and "$b$" reversed.

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Another alternative: $f(x)=\frac{2}{x}$ is convex for $x>0$. Therefore $\frac{2}{a}+\frac{2}{b}\geq 2\frac{2}{\frac{a+b}{2}}$ or equivalently $\frac{2}{a}+\frac{2}{b}\geq \frac{8}{a+b}>\frac{4}{a+b}$

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Let's assume that: $$ \frac{2}{a}+\frac{2}{b}=\frac{4}{a+b} $$ $\text{ Where }a,b>0 \text { and } a,b \in R$ $$ \begin{aligned} &\begin{array}{l} \Rightarrow \frac{2(a+b)}{a b}=\frac{4}{a+b} \\ \Rightarrow \quad(a+b)^{2}=2 a b \end{array}\\ &\Rightarrow a^{2}+b^{2}=0\\ &\Rightarrow \quad a^{2}=-b^{2} \end{aligned} $$

Now it implies that square of $a$ is negative of $b$'s square. But it can't be possible 'cause both $a,b>0$ and it is elementary knowledge that square of any positive number can't be negative.

Now as every step of our above solution is correct, hence our assumption was wrong.

Therefore' $$ \frac{2}{a}+\frac{2}{b}\not=\frac{4}{a+b} \text{ Q.E.D } $$

EXTRA: Actually $$ \frac{2}{a}+\frac{2}{b}\not=\frac{4}{a+b} $$ is true for every real number$a$ and $b$ such that $a\not=0 \text{ and } b\not=0$.(why?)

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