# Symmetric Inequality Problem

I am struggling with understanding https://nrich.maths.org/251.

To restate the problem,

If $x$, $y$ and $z$ are real numbers such that: $\begin{cases}x+y+z=5 \\ xy+yz+zx=3\end{cases}$ , What is the largest value that any one of these numbers can have?

In particular, I do not understand the first solution given, and while the second I am getting a grip with (creates a quadratic uses the discriminant inequality since $x$, $y$ and $z$ are real numbers), would like to ask whether any classical inequalities can be used here, as I would be personally more satisfied with this.

The problem I had with applying inequalities I knew was that $x$, $y$ and $z$ could be any real numbers, not just positive.

Any help/discussion would be much appreciated! Note by Arthur Conmy
2 years, 6 months ago

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The link u have referred is not opening, it says the page not found.However, u can see my solution to this problem : https://brilliant.org/problems/almost-vietas/#!/solution-comments/171217/

- 2 years, 6 months ago

Fixed btw

- 2 years, 6 months ago

yup

- 2 years, 6 months ago

- 2 years, 6 months ago

actually u have put a dot by mistake in front of 251

- 2 years, 6 months ago

okay ! great

- 2 years, 6 months ago

Hey buddy is the answer $\frac{13}{2}$?

- 2 years, 6 months ago

i m getting $\frac{13}{3}$ as maximum value and -1 as minimum value

- 2 years, 6 months ago

Yep same here! What was your method?

- 2 years, 6 months ago

but u said u are getting $\frac{13}{2}$ . I used the same method which i referred to in the link i gave.

- 2 years, 6 months ago

Oops! Sorry I am getting $\frac{13}{3}$. I actually did not check the link out. Let me check it now

- 2 years, 6 months ago

- 2 years, 6 months ago

Cauchy Schwarz inequality

- 2 years, 6 months ago

same method as mine then? or in a different way?

- 2 years, 6 months ago