<p>You have </p>

<p>$$ \begin{align} u(\lambda a + (1 - \lambda) b, \lambda b + (1 - \lambda)a) &amp;= \sqrt{\lambda a + (1 - \lambda)b} + \sqrt{\lambda b + (1 - \lambda)a} \\ 
&amp;\geq \lambda\sqrt{a} + (1 - \lambda)\sqrt{b} + \lambda\sqrt{b} + (1 - \lambda)\sqrt{a} \\ &amp;=  \sqrt{a} + \sqrt{b} \\ &amp;= u(a, b) \end{align}$$</p>

<p>The inequality in the second line follows from Jensen's inequality for concave functions. I don't think you need to prove Jensen but you can even do that moderately easy with the AM-GM inequality if you want to. </p>

CC BY-SA 3.0
2018-02-12T20:00:52.070Z
2018-02-12T20:00:52.070Z
Taufi
<p>I need some help with a proof.</p>

<blockquote>
  <p>Assume η = 2 and there are just two goods.  Verify that the following utility function exhibits Love for Variety tastes, show that:</p>
  
  <p>u(λa + [1 − λ]b, λb + [1 − λ]a) > u(a, b) for 1 > λ > 0 </p>
  
  <p>Additionally, I am given the utility function:
  $$
\\u(q_1,q_2, ...q_n) = \sum_1^N \sqrt[η]{q_i}\,
$$</p>
</blockquote>

<p>In the lead up to the question, I am given a different problem to prove that should help me solve the above one.  </p>

<p>It uses the same utility function, and I have to prove that the consumer wants a variety of two firms, rather than just one firm, I think I figured it out but I don't know how to apply what I did in the above proof, I have to prove that sqrt(x) + sqrt(y) > sqrt(x + y):</p>

<p>$$
\sqrt{x} + \sqrt{y} &gt; \sqrt{x+y},
$$
$$
(\sqrt{x} + \sqrt{y})^2 &gt; (\sqrt{x + y})^2
$$
$$
2\sqrt{xy} + x + y &gt; x + y
$$
$$
2\sqrt{xy} &gt; 0
$$</p>

<p>Any help is sincerely appreciated.</p>

CC BY-SA 3.0
2018-02-10T21:39:55.757Z
What are $a$, $b$, $q_i$, $x$ and $y$? What do you mean by "a variety of two firms"?
Herr K.
CC BY-SA 4.0
2019-05-26T17:09:25.557Z
@colorado I disagree with your method of proof for $\sqrt x + \sqrt y > \sqrt{x + y}$. You can't fiddle with a relation - you should rather modify each side separately. Btw, questions like these would be received better on [math.stackexchange.com](https://math.stackexchange.com/).
ahorn
<industrial-organisation><proof>
Help with Monopolistic Competition Proof, Prove Love for Variety
colorado
# Help with Monopolistic Competition Proof, Prove Love for Variety

(Asked by: username_0 on 2018-02-10)

I need some help with a proof.

> Assume η = 2 and there are just two goods. Verify that the following utility function exhibits Love for Variety tastes, show that:
>
> u(λa + [1 − λ]b, λb + [1 − λ]a) > u(a, b) for 1 > λ > 0
>
> Additionally, I am given the utility function:
> $$
> \\u(q\_1,q\_2, ...q\_n) = \sum\_1^N \sqrt[η]{q\_i}\,
> $$

In the lead up to the question, I am given a different problem to prove that should help me solve the above one.

It uses the same utility function, and I have to prove that the consumer wants a variety of two firms, rather than just one firm, I think I figured it out but I don't know how to apply what I did in the above proof, I have to prove that sqrt(x) + sqrt(y) > sqrt(x + y):

$$
\sqrt{x} + \sqrt{y} > \sqrt{x+y},
$$
$$
(\sqrt{x} + \sqrt{y})^2 > (\sqrt{x + y})^2
$$
$$
2\sqrt{xy} + x + y > x + y
$$
$$
2\sqrt{xy} > 0
$$

Any help is sincerely appreciated.

**Tags:** industrial-organisation, proof

> **Comments:**
> **username_1**: What are $a$, $b$, $q_i$, $x$ and $y$? What do you mean by "a variety of two firms"?  
> **username_2**: @username_0 I disagree with your method of proof for $\sqrt x + \sqrt y > \sqrt{x + y}$. You can't fiddle with a relation - you should rather modify each side separately. Btw, questions like these would be received better on [math.stackexchange.com](https://math.stackexchange.com/).  

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## 1 answers

### Answer by username_3 on 2018-02-12. Score: 1

You have

$$ \begin{align} u(\lambda a + (1 - \lambda) b, \lambda b + (1 - \lambda)a) &= \sqrt{\lambda a + (1 - \lambda)b} + \sqrt{\lambda b + (1 - \lambda)a} \\
&\geq \lambda\sqrt{a} + (1 - \lambda)\sqrt{b} + \lambda\sqrt{b} + (1 - \lambda)\sqrt{a} \\ &= \sqrt{a} + \sqrt{b} \\ &= u(a, b) \end{align}$$

The inequality in the second line follows from Jensen's inequality for concave functions. I don't think you need to prove Jensen but you can even do that moderately easy with the AM-GM inequality if you want to.


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