1. ## Inequalities

1.

Prove that

$a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab$

holds for all real numbers a, b and c

(>= means greater than or equal to)

2.

Prove that if real numbers a, b and c satisfy

$a + b + c> 0$

$ab +ac +bc > 0$

$abc > 0$

then each of a, b, and c is positive

Thanks for the help guys ^_^

2. Originally Posted by JintoLyn
1.

Prove that

$a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab$

holds for all real numbers a, b and c

(>= means greater than or equal to)

2.

Prove that if real numbers a, b and c satisfy

$a + b + c> 0$

$ab +ac +bc > 0$

$abc > 0$

then each of a, b, and c is positive

Thanks for the help guys ^_^
Here's the first. Consider

$
\frac{(a^2-b^2)^2}{2} + \frac{(b^2-c^2)^2}{2} + \frac{(c^2-a^2)^2}{2} \ge 0
$

and

$
(a^2-bc)^2 + (b^2-a c)^2+ (c^2-ab)^2 \ge 0
$
.

3. Another way for the first one: AM-GM inequality:

$
a^4 + b^4 + c^4 = \frac{a^4 + a^4+b^4+c^4}{4} + \frac{a^4+b^4+b^4+c^4}{4} + \frac{a^4+b^4+c^4+c^4}{4}$
$
\ge a^2bc+ab^2c+abc^2
$

For the second one, consider the polynomial:
$
(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc =$

$
= x^3-\alpha x^2 + \beta x - \gamma
$

Where $\alpha > 0\,\,,\,\, \beta > 0\,\,,\,\, \gamma > 0$

Assume one of the roots, say $c$, is negative. Put $x=c$ in the above equality and you'll get 0 < 0, which is absurd.

4. I believe both can be solved by contradiction.

1.

Assume the opposite is true:
$a^4 + b^4 + c^4 < a^2bc + b^2ac + c^2ab$ for all integers.

or for simplicity

$\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} < a + b + c$

we can instantly see by inspection that any integer $a = b = c$ does not work since $a + b + c < a + b + c$ is false.

2.

From $abc>0$ and assuming that inequality is correct you can instantly get $a>0, b>0, c>0$.

If you start off by assuming one or more variable is less than zero and also assume this set of equations work for all positive or negative integers (depending upon which variable you assume is less than or greater than zero) you will run into contradictions

EDIT: I didn't see the first two replies prior to posting this.. you guys are too quick for me

5. Originally Posted by JintoLyn
1.

Prove that

$a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab$

holds for all real numbers a, b and c

(>= means greater than or equal to)

2.

Prove that if real numbers a, b and c satisfy

$a + b + c> 0$

$ab +ac +bc > 0$

$abc > 0$

then each of a, b, and c is positive

Thanks for the help guys ^_^
Here's the second. From the third equation $abc > 0$ so $a \ne 0, b \ne 0 \; \text{and}\; c \ne 0$. Also due to the sign of the third equation either all are positive or two of the three are negative. We will show only the first case is true. Assume that $b$ and $c$ are negative. Let $b = - \bar{b}$ and $c = - \bar{c}$ so $\bar{b} > 0$ and $\bar{c} > 0$.

Thus $a - \bar{b} - \bar{c} > 0\;\; \Rightarrow\;\; a > \bar{b} + \bar{c} > 0$

and

$
- a \bar{b} - a \bar{c} + \bar{b}\, \bar{c} > 0\;\; \Rightarrow\;\; \bar{b}\, \bar{c} > a \bar{b} + a \bar{c} > 0
$
.

Multiplying the inequalities gives

$
a\, \bar{b} \, \bar{c} > a\left(\bar{b}+\bar{c}\right)^2\;\; \Rightarrow\;\; 0 > \bar{b}^2 + \bar{b}\,\bar{c} + \bar{c}^2
$

which leads to a contradiction since $\bar{b}, \bar{c} > 0$.

Too slow but with details

6. Originally Posted by JintoLyn
1.

Prove that

$a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab$

holds for all real numbers a, b and c

(>= means greater than or equal to)

$x_1\ \ge\ x_2,\ y_1\ \ge\ y_2\ \Rightarrow\ x_1y_1+x_2y_2\ \ge\ x_1y_2+x_2y_1$

This is because $(x_1-x_2)(y_1-y_2)\ \ge\ 0$

$x_1(y_1-y_2)-x_2(y_1-y_2)\ \ge\ 0$

$x_1y_1-x_1y_2-x_2y_1+x_2y_2=x_1y_1+x_2y_2-(x_1y_2+x_2y_1)\ \ge\ 0$

Then, if $a\ \ge\ b,\ b\ \ge\ c$

$ab+bc\ \ge\ ac+bb$

$a^2bc+b^2ac+c^2ab=a(abc)+b(abc)+c(abc)=(a+b+c)abc$

$a^4+b^4+c^4=a^3a+b^3b+c^3c$

$\ge\ a^3a+b^2bc+c^2cb$

$\ge\ a^2ab+bbac+c^2cb$

$\ge\ bbac+aba^2+bcc^2$ by just rewriting the previous line

$\ge\ b(abc)+abac+bcac$

$\ge\ b(abc)+a(abc)+c(abc)$

Hence

$a^4+b^4+c^4\ \ge\ a^2bc+b^2ac+c^2ab$

7. Originally Posted by JintoLyn
2.

Prove that if real numbers a, b and c satisfy

$a + b + c> 0$

$ab +ac +bc > 0$

$abc > 0$

then each of a, b, and c is positive

Thanks for the help guys ^_^
$a+b+c\ >\ 0$

$ab+ac+bc\ >\ 0$

factorising the 2nd line line in 3 ways we get

$\color{blue}a(b+c)+bc\ >0$

$\color{blue}ab+(a+b)c\ >0$

$\color{blue}ac+(a+c)b\ >0$

$abc\ >0$

$a+b+c\ >0\ \Rightarrow\ b+c\ >-a$

$a(b+c)+bc\ >0\ \Rightarrow\ bc>-a(b+c)\ >-a(-a)\ >a^2$

Hence $bc>0$

It follows from the other two blue inequalities that $ab>0,\ bc>0$

Therefore since abc>0, each of a, b, c are positive