Prove that the following inequality holds for all real numbers a, b and c.?
a^4 + b^4 + c^4 is greater or equal to (a^2)(b)(c) + (b^2)(a)(c) + (c^2)(a)(b)
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a^4+b^4+c^4=>(a^2)(b)(c)+(b^2)(a)(b)+(c^2)(a)(b)
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6 years ago
By the Cauchy-Schwarz inequality,
(a^2)(bc) + (b^2)(ac) + (c^2)(ab)<= sqrt((a^4+b^4+c^4)*((bc)^2+(ac)^2+(ab)^2)). But (bc)^2=(b^2)*(c^2)<=(b^4+c^4)/2, (ac)^2=(a^2)*(c^2)<=(a^4+c^4)/2, (ab)^2=(a^2)*(b^2)<=(a^4+b^4)/2. Adding up the above three inequalities, we conclude that (bc)^2+(ac)^2+(ab)^2<=a^4+b^4+c^4. Thus, (a^2)(b)(c) + (b^2)(a)(c) + (c^2)(a)(b) <=sqrt((a^4+b^4+c^4)^2) =a^4+b^4+c^4 as desired.
(a^2)(bc) + (b^2)(ac) + (c^2)(ab)<= sqrt((a^4+b^4+c^4)*((bc)^2+(ac)^2+(ab)^2)). But (bc)^2=(b^2)*(c^2)<=(b^4+c^4)/2, (ac)^2=(a^2)*(c^2)<=(a^4+c^4)/2, (ab)^2=(a^2)*(b^2)<=(a^4+b^4)/2. Adding up the above three inequalities, we conclude that (bc)^2+(ac)^2+(ab)^2<=a^4+b^4+c^4. Thus, (a^2)(b)(c) + (b^2)(a)(c) + (c^2)(a)(b) <=sqrt((a^4+b^4+c^4)^2) =a^4+b^4+c^4 as desired.
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