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sqing

#1 h Wednesday at 12:49 PM

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Let $a,b,c\geq 2 .$ Prove that
$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$ $(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$ $(a^2-1)(b-1)(c^2-1) -2abc\geq -7$ $(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$ $(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$ $(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$

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sqing

#2 h Wednesday at 1:03 PM

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Let $a,b,c\geq 2 .$ Prove that
$(a-1)(b^3-1)(c -1) -\frac{7}{4}abc\geq -7$ $(a^3-1)(b-1)(c^3-1) -\frac{49}{4}abc\geq -49$

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#3 h Wednesday at 1:25 PM

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sqing wrote:

Let $a,b,c\geq 2 .$ Prove that
$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$ $(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$ $(a^2-1)(b-1)(c^2-1) -2abc\geq -7$ $(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$ $(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$ $(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$

$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$ $(a-1)(b-1)(c-1) \geq \frac{1}{4}abc-1$ expand : $3abc+4\sum a\geq 4\sum ab$ put : $a=x+2$ $b=y+2$ $c=z+2$ $x,y,z\geq 0$ $3(x+2)(y+2)(z+2)+4\sum (x+2)\geq 4\sum (x+2)(y+2)$ expand : $xyz+2\sum xy\geq 0$

This post has been edited 3 times. Last edited by SunnyEvan, Wednesday at 1:35 PM

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sqing

#4 h Wednesday at 2:21 PM

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Thanks.
$(a-1)(b-1)(c-1)- \frac{1}{4}abc \geq-1$ That's right.

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sqing

#5 h Wednesday at 2:34 PM

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Let $a,b,c\geq 2 .$ Prove that
$(a-1)(b^2-1)(c^3-1)- \frac{21}{4}abc\geq -21$ $(a-1)(b^2-1)(c^3-2)- \frac{9 }{2}abc \geq -18$ $(a-1)(b^2-1)(c^3-3)- \frac{15}{4}abc \geq -15$ $(a-1)(b^2-2)(c^3-2)- 3abc\geq -12$

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sqing

#6 h Yesterday at 2:35 AM

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sqing wrote:

Let $a,b,c\geq 2 .$ Prove that
$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$

sqing wrote:

Let $a,b,c\geq 2 .$ Prove that
$(a-1)(b^2-2)(c-1) - \frac{1}{2}abc\geq -2$

Solution of no_room_for_error:
$\begin{aligned} (a-1)(b^2-2)(c-1) &\geq 2(a-1)(b-1)(c-1)\\ &=\frac{1}{2}abc-2+\frac{3}{2}(a-2)(b-2)(c-2)+(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)\\ &\geq \frac{1}{2}abc-2\\ \end{aligned}$ Equality occurs when $b=2$ and $(c-2)(a-2)=0$ .

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