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

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

• what have You tried so far and which part confuses You? – Rezha Adrian Tanuharja Apr 7 at 1:01
• Please typeset your question (in MathJax) and not post pictures, which cannot be searched. – 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, $$a,b>0$$.

So,

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

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

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

Expanding and simplifying:

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

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.

• can i proof by contraposition ?? – user13240088 Apr 7 at 1:24
• @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. – 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 $$n< n+k, k so $$\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.

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 $$\frac{2}{a}+\frac{2}{b}\geq \frac{8}{a+b}>\frac{4}{a+b}$$

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?)