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Which of these terms is greater ?

$2x-6y+1$ or $1$

if $x^4 + 3y^2=0$

According to the text they are equal ?How is that ?

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  • $\begingroup$ It boils down to comparing $2x-6y$ and $0$, which further reduces to comparing $x$ and $3y$. $\endgroup$ – Karthik C Jul 13 '12 at 5:36
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But since $x^2$ and $3y^2$ are positive, (assuming $x,y$ real) we have that $x^2+3y^2=0$ forces $x=y=0$ so the text is correct!

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What number system are $x,y$ in? If $x,y\in\mathbb{R}$, then it is simple, since $x^4$, $3y^2$ $\ge 0$, we must have that $x,y=0$, hence $2x-6y+1=1$.

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If sum of squares is $0$ then each term of the series is $0$ individually (if they belong to $\Bbb R$) $\implies$ $x=0,y=0$ in your problem which gives $2x-6y+1=1$.

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we know that for positive terms Arithmetic Mean $\geq$ Geometric Mean

i.e. $$(x^4 + y^2 + y ^2 + y^2)/4 \geq \sqrt[4]{(x^4)(y^2)(y^2)(y^2)}$$

i.e. $(x^4 + 3y^2)/4 \ge (x)(y)\sqrt{y}$

i.e. $(x)(y)\sqrt{y} \le (x^4 + 3y^2)/4$

i.e. $(x)(y)\sqrt{y} \le 0$

for $\sqrt{y}$ to be real $y \ge 0$

i.e. $y \ge 0$ ------- (1)

i.e. $x \le 0$ ------- (2)

we know that for positive terms Arithmetic Mean $\geq$ Harmonic Mean

i.e. $$(x^4 + y^2 + y^2 + y^2)/4 \geq {4/(1/x^4 + 1/y^2 + 1/y^2 + 1/y^2)}$$

i.e. $$(x^4 + 3y^2)(1/x^4 + 3/y^2) \geq {(4)(4)}$$

i.e. $$(1 + 3(y/x^2)^2 + 3(x^2/y)^2 + 9) \geq {16}$$

i.e. $$3(y^2/x^4) + 3(x^4/y^2) \geq {6}$$

i.e. $$(y^2/x^4) + (x^4/y^2) \geq {2}$$

i.e. $$(y^4 + x^8) \geq {2(y^2)(x^4)}$$

i.e. $$(y^2 - x^4)^2 \geq {0}$$

i.e. $$(x^4 - y^2)^2 \geq {0}$$

consider $$x^4 \geq {y^2}$$

i.e. $$(x^4 - y^2) \geq {0}$$

i.e. $$(x^2 - y)(x^2 + y) \geq {0}$$

i.e. $$(x^2 - y)(x^2 + y) \geq {0}$$

i.e. $$(y - x^2)(y + x^2) \leq {0}$$

i.e. ${-x^2} \leq y \leq {x^2}$ ------- (3)

The three inequalities listed above will hold concurrently only when $x = y = 0$

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  • 5
    $\begingroup$ Please use $LATEX$ $\endgroup$ – Aang Jul 13 '12 at 6:57
  • $\begingroup$ x^4 + 3y^2 = 0 implies each term x^4 and y^2 are individually equal to 0 $\endgroup$ – HOLYBIBLETHE Jul 13 '12 at 6:58
  • $\begingroup$ NOTE : for sqrt(y) to be real y ge 0 $\endgroup$ – HOLYBIBLETHE Jul 13 '12 at 14:25

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