How to show this inequality a14−a13+a8−a5+a2−a+1&gt;0

Notice that quadratic x2−x+1&gt;0 for all x. Now write: (a−1)(a13−1)⏟≥0+a2(a6−a3+1⏟&gt;0)&gt;0

Notice it a bunch of +aeven and −aodd in descending order of powers so..... Case 1: a≤0 then aeven≥0and −aodd≥0 so a14−a13+a8−a5+a2−a+1=|a|14+|a|13+|a|8+|a|5+|a|2+|a|+1≥1. Case 2: a≥1then abig≥asmall so abig−asmall&gt;≥0 and (a14−a13)+(a8−a5)+(a2−a)+1≥1.

Notice that quadratic x2−x+1&gt;0 for all x. Now write: (a−1)(a13−1)⏟≥0+a2(a6−a3+1⏟&gt;0)&gt;0

Notice it a bunch of +aeven and −aodd in descending order of powers so..... Case 1: a≤0 then aeven≥0and −aodd≥0 so a14−a13+a8−a5+a2−a+1=|a|14+|a|13+|a|8+|a|5+|a|2+|a|+1≥1. Case 2: a≥1then abig≥asmall so abig−asmall&gt;≥0 and (a14−a13)+(a8−a5)+(a2−a)+1≥1.

Best solutions to - "How to show this inequality a14−a13+a8−a5+a2−a+1>0"

High School
Algebra
Algebra I
Quadratic function and equation

Alvin Pugh

Answered question

2022-01-31

How to show this inequality

a^{14} - a^{13} + a^{8} - a^{5} + a^{2} - a + 1 > 0

Answer & Explanation

shimmertulipsog

Beginner2022-02-01Added 10 answers

Notice that quadratic

x^{2} - x + 1 > 0

for all x. Now write:

\underset{\geq 0}{\underset{⏟}{(a - 1) (a^{13} - 1)}} + a^{2} (\underset{> 0}{\underset{⏟}{a^{6} - a^{3} + 1}}) > 0

ataill0k

Beginner2022-02-02Added 18 answers

Notice it a bunch of

+ a^{even} and - a^{odd}

in descending order of powers so.....
Case 1:

a \leq 0 then a^{e v e n} \geq 0 and - a^{o d d} \geq 0

a^{14} - a^{13} + a^{8} - a^{5} + a^{2} - a + 1 = {| a |}^{14} + {| a |}^{13} + {| a |}^{8} + {| a |}^{5} + {| a |}^{2} + | a | + 1 \geq 1

.
Case 2:

a \geq 1 then a^{b i g} \geq a^{s m a l l} so a^{b i g} - a^{s m a l l} > \geq 0 and (a^{14} - a^{13}) + (a^{8} - a^{5}) + (a^{2} - a) + 1 \geq 1

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