An inequality withx,y,z≥−1 and x+y+z=1

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Ellie Horne · Accepted Answer

Step 1I think it means thatxyz≠0Let x=43a−1, y=43b−1 and z=43c−1Thus, a, b and c are non-negatives, a+b+c=3 and we need to prove that:8∑cyc(43a−1)2∏cyc(43a−1)+1+9∏cyc(43a−1)2≥30∏cyc(43a−1)or8∏cyc(4a−3)∑cyc(4a−3)2+243+3∏cyc(4a−3)2≥270∏cyc(4a−3)or8∏cyc(4a−3)(16(a2+b2+c2)−45)+243+3∏cyc(4a−3)2≥270∏cyc(4a−3).Now, let a+b+c=3u, ab+ac+bc=3v2, abc=w3 and u2=tv2Thus, t≥1 and we need to prove that:8(64w3−144uv2+81u3)(33u2−32v2)u+81u6+(64w3−144uv2+81u3)2≥90(64w3−144uv2+81u3)u3or16w6+4u(21u2−34v2)w3+9u2(3u2−5v2)2≥0.We see that if 21u2−34v2≥0 so the inequality is true.Thus, it&#039;s enough assume that 1≤t≤3421.In another hand, if4u2(21u2−34v2)2−144u2(3u2−5v2)2&amp;lt;0 so the inequality is true and it&#039;s enough to prove our inequality for4u2(21u2−34v2)2−144u2(3u2−5v2)2≥0or4u2(3u2−4v2)(39u2−64v2)≥0,which with 1≤t≤3421 gives 1≤t≤43 and it&#039;s enough to prove that:w3≤−2u(21u2−34v2)−4u2(3u2−4v2)(39u2−64v2)16orw3≤34uv2−21u3−u(3u2−4v2)(39u2−64v2)8.But∏cyc(a−b)2=27(3u2v4−4v6−4u3w3+6uv2w3−w6)≥0givesw3≤3uv2−2u3+2(u2−v2)3and it&#039;s enough to prove that:3uv2−2u3+2(u2−v2)3≤34uv2−21u3−u(3u2−4v2)(39u2−64v2)8or10uv2−5u3−16(u2−v2)3≥u(3u2−4v2)(39u2−64v2)or10−5t−16(t−1)3t≥(3t−4)(39t−64).We&#039;ll prove that10−5t−16(t−1)3t≥0.Indeed, it&#039;s25t(2−t)2≥256(t−1)3or231t3−668t2+668t−256≤0or(3t−4)(77t2−120t+62)+2t−8≤0,which is true because 602−77⋅62&amp;lt;0Thus, it&#039;s enough to prove that:(10−5t−16(t−1)3t)2≥(3t−4)(39t−64)or64(t−1)3t−40(2−t)(t−1)3t≥23t2−62t+39or since 41t2−89t+64&amp;gt;0,(41t2−89t+64)2≥1600(2−t)2t(t−1)or(9t2+39t−64)2≥0and we are done!It&#039;s interesting that since 1≤−13+5176≤34,, the equality occurs in seven different points!

An inequality with \(\displaystyle{x},{y},{z}\ge-{1}\) and \(\displaystyle{x}+{y}+{z}={1}\)

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