Inequality with real numbers

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Inequality with real numbers

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#1 h Yesterday at 6:48 AM • 1 Y

Y by arqady

Let a,b,c are real numbers. Prove that (a^3+b^3+c^3+3abc)^4+(a+b+c)^3(a+b-c)^3(-a+b+c)^3(a-b+c)^3>=0

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#3 h Yesterday at 10:18 AM

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$(a^3+b^3+c^3+3abc)^4+(a+b+c)^3(a+b-c)^3(-a+b+c)^3(a-b+c)^3\geq 0$ ==> $(a^3+b^3+c^3+3abc)^4+[(a+b+c)(a+b-c)(b+c-a)(c+a-b)]^3\geq 0$ because : $(a+b+c)(a+b-c)(b+c-a)(c+a-b)=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$ so $(a^3+b^3+c^3+3abc)^4+(a+b+c)^3(a+b-c)^3(-a+b+c)^3(a-b+c)^3\geq 0$ <===> $(a^3+b^3+c^3+3abc)^4 \geq ((a^4+b^4+c^4)-2(a^2b^2+b^2c^2+c^2a^2))^3$ Substituting a,b,c by |a|,|b|,|c|, respectively, the right side of the inequality remains unchanged, while the left side either remains unchanged or increases. Therefore, it suffices to prove the inequality only for a,b,c ≥ 0

When $(a^4+b^4+c^4)-2(a^2b^2+b^2c^2+c^2a^2) \leq 0$ , the inequality clearly holds.

When $(a^4+b^4+c^4)-2(a^2b^2+b^2c^2+c^2a^2) \geq 0$ , the inequality become: $(a^3+b^3+c^3+3abc)^4 \geq |(a^4+b^4+c^4)-2(a^2b^2+b^2c^2+c^2a^2)|^3$ The conditions for the establishment of an equal sign are :
1. any two variables are 0 (another variable takes a random value).
2. one variable is zero, and the other two variables are opposite numbers to each other.

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#4 h Yesterday at 10:20 AM

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I'm not sure

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arqady

#5 h Yesterday at 4:19 PM

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JK1603JK wrote:

Let a,b,c are real numbers. Prove that $(a^3+b^3+c^3+3abc)^4+(a+b+c)^3(a+b-c)^3(-a+b+c)^3(a-b+c)^3\geq0$

Let $a+b+c=3u$ , $ab+ac+bc=3v^2$ and $abc=w^3$ .
Thus, we need to prove that:
$3(9u^3-9uv^2+2w^3)^4+u^3(-27u^3+36uv^2-8w^3)^3\geq0$ or
$(9uv^2-2w^3)^2(12w^6+4u(22u^2-27v^2)w^3+9u^2(18u^4-44u^2v^2+27v^4))\geq0,$ for which it's enough to prove that:
$4u^2(22u^2-27v^2)^2-108u^2(18u^4-44u^2v^2+27v^4)\leq0$ or $u^6\geq0.$

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