Inequalities

May 2010
2
0
1.

Prove that

\(\displaystyle a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab\)

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


\(\displaystyle a + b + c> 0 \)

\(\displaystyle ab +ac +bc > 0 \)

\(\displaystyle abc > 0\)

then each of a, b, and c is positive


Thanks for the help guys ^_^
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
1.

Prove that

\(\displaystyle a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab\)

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


\(\displaystyle a + b + c> 0 \)

\(\displaystyle ab +ac +bc > 0 \)

\(\displaystyle abc > 0\)

then each of a, b, and c is positive


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

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

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

Expand and add and your inequality will follow.
 
Nov 2009
106
105
Another way for the first one: AM-GM inequality:

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

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

Where \(\displaystyle \alpha > 0\,\,,\,\, \beta > 0\,\,,\,\, \gamma > 0\)

Assume one of the roots, say \(\displaystyle c\), is negative. Put \(\displaystyle x=c\) in the above equality and you'll get 0 < 0, which is absurd.
 
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May 2010
17
2
Cleveland,OH/Madison,WI/Queens,NY
I believe both can be solved by contradiction.

1.

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

or for simplicity

\(\displaystyle \frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} < a + b + c\)

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

2.

From \(\displaystyle abc>0\) and assuming that inequality is correct you can instantly get \(\displaystyle a>0, b>0, c>0\).

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 :)
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
1.

Prove that

\(\displaystyle a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab\)

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


\(\displaystyle a + b + c> 0 \)

\(\displaystyle ab +ac +bc > 0 \)

\(\displaystyle abc > 0\)

then each of a, b, and c is positive


Thanks for the help guys ^_^
Here's the second. From the third equation \(\displaystyle abc > 0\) so \(\displaystyle a \ne 0, b \ne 0 \; \text{and}\; c \ne 0\). 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 \(\displaystyle b\) and \(\displaystyle c\) are negative. Let \(\displaystyle b = - \bar{b}\) and \(\displaystyle c = - \bar{c}\) so \(\displaystyle \bar{b} > 0\) and \(\displaystyle \bar{c} > 0\).

Thus \(\displaystyle a - \bar{b} - \bar{c} > 0\;\; \Rightarrow\;\; a > \bar{b} + \bar{c} > 0\)

and

\(\displaystyle
- a \bar{b} - a \bar{c} + \bar{b}\, \bar{c} > 0\;\; \Rightarrow\;\; \bar{b}\, \bar{c} > a \bar{b} + a \bar{c} > 0
\).

Multiplying the inequalities gives

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

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

Too slow but with details (Rofl)
 
Dec 2009
3,120
1,342
1.

Prove that

\(\displaystyle a^4 + b^4 + c^4 >= a^2bc+b^2ac+c^2ab\)

holds for all real numbers a, b and c

(>= means greater than or equal to)
\(\displaystyle x_1\ \ge\ x_2,\ y_1\ \ge\ y_2\ \Rightarrow\ x_1y_1+x_2y_2\ \ge\ x_1y_2+x_2y_1\)

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

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

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

Then, if \(\displaystyle a\ \ge\ b,\ b\ \ge\ c\)

\(\displaystyle ab+bc\ \ge\ ac+bb\)

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

\(\displaystyle a^4+b^4+c^4=a^3a+b^3b+c^3c\)

\(\displaystyle \ge\ a^3a+b^2bc+c^2cb\)

\(\displaystyle \ge\ a^2ab+bbac+c^2cb\)

\(\displaystyle \ge\ bbac+aba^2+bcc^2\) by just rewriting the previous line

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

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

Hence

\(\displaystyle a^4+b^4+c^4\ \ge\ a^2bc+b^2ac+c^2ab\)
 
Dec 2009
3,120
1,342
2.

Prove that if real numbers a, b and c satisfy


\(\displaystyle a + b + c> 0 \)

\(\displaystyle ab +ac +bc > 0 \)

\(\displaystyle abc > 0\)

then each of a, b, and c is positive


Thanks for the help guys ^_^
\(\displaystyle a+b+c\ >\ 0\)

\(\displaystyle ab+ac+bc\ >\ 0\)

factorising the 2nd line line in 3 ways we get

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

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

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

\(\displaystyle abc\ >0\)

\(\displaystyle a+b+c\ >0\ \Rightarrow\ b+c\ >-a\)

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

Hence \(\displaystyle bc>0\)

It follows from the other two blue inequalities that \(\displaystyle ab>0,\ bc>0\)

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