Find the maximum value of x^3+2y

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Find the maximum value of x^3+2y

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Source: 2021 Turkey JBMO TST P4

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#1 h May 23, 2021, 7:56 PM • 4 Y

Y by centslordm, teomihai, Mango247, ehuseyinyigit

Let $x,y,z$ be real numbers such that $\left|\dfrac yz-xz\right|\leq 1\text{ and }\left|yz+\dfrac xz\right|\leq 1$ Find the maximum value of the expression $x^3+2y$

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#2 h May 23, 2021, 7:57 PM • 4 Y

Y by centslordm, chess12500, wizixez, Nurefsan17

Answer: $2$ .
Equality Case: $(x,y,z)=(0,1,1)$ or $(x,y,z)=(0,1,-1)$
Proof:
$\left|\dfrac yz-xz\right|\leq 1\Rightarrow \dfrac {y^2}{z^2}+x^2z^2-2xy\leq 1$ .
$\left|\dfrac xz+yz\right|\leq 1\Rightarrow \dfrac {x^2}{z^2}+y^2z^2+2xy\leq 1$ .
Then $\dfrac {y^2}{z^2}+x^2z^2+\dfrac {x^2}{z^2}+y^2z^2\leq 2\Rightarrow (x^2+y^2)(z^2+\dfrac 1{z^2})\leq 2$ .

By AM-GM $z^2+\dfrac 1{z^2}\ge 2$ . Hence, $x^2+y^2\leq 1$ .

Let $f(x,y)=x^3+2y$ . We are searching for the maximum value of this function while $x^2+y^2\leq 1$ .

Obviously $f(x,y)\leq f(|x|,|y|)$ . So, we can assume $x,y\ge 0$ .

$f(x,y)$ is increasing while $x$ is increasing. So, we can assume $x^2+y^2=1$ .

$x^3+2y\leq 2\Leftrightarrow x^3+2(y^3+x^2y)\leq 2\Leftrightarrow (x^3+2y^3+2x^2y)^2\leq 4(x^2+y^2)^3\Leftrightarrow$
$x^6+4y^6+4x^4y^2+4x^3y^3+4x^5y+8y^4x^2\leq 4x^6+4y^6+12x^4y^2+12y^4x^2\Leftrightarrow$
$4x^3y^3+4x^5y\leq 3x^6+8x^4y^2+4y^4x^2\Leftrightarrow 4xy^3+4x^3y\leq 3x^4+8x^2y^2+4y^4$ .

By AM-GM $x^4+x^4+x^4+y^4\ge 4x^3y$ . Then it suffices to prove that $8x^2y^2+3y^4\ge 4xy^3$ .

$8x^2y^2+3y^4\ge 4xy^3\Leftrightarrow 8x^2+3y^2\ge 4xy\Leftrightarrow 2(x-y)^2+6x^2+y^2\ge 0$ , which is correct.

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#3 h Jun 10, 2021, 5:37 AM • 2 Y

Y by teomihai, ehuseyinyigit

BarisKoyuncu wrote:

Answer: $2$ .
Equality Case: $(x,y,z)=(0,1,1)$ or $(x,y,z)=(0,1,-1)$
Proof:
$\left|\dfrac yz-xz\right|\leq 1\Rightarrow \dfrac {y^2}{z^2}+x^2z^2-2xy\leq 1$ .
$\left|\dfrac xz+yz\right|\leq 1\Rightarrow \dfrac {x^2}{z^2}+y^2z^2+2xy\leq 1$ .
Then $\dfrac {y^2}{z^2}+x^2z^2+\dfrac {x^2}{z^2}+y^2z^2\leq 2\Rightarrow (x^2+y^2)(z^2+\dfrac 1{z^2})\leq 2$ .

By AM-GM $z^2+\dfrac 1{z^2}\ge 2$ . Hence, $x^2+y^2\leq 1$ .

Let $f(x,y)=x^3+2y$ . We are searching for the maximum value of this function while $x^2+y^2\leq 1$ .

Obviously $f(x,y)\leq f(|x|,|y|)$ . So, we can assume $x,y\ge 0$ .

hence another continuation below
$x^3+2y\le x^2+2\sqrt{1-x^2}\le x^2+1+1-x^2=2$ done.

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sqing

#4 h Jun 10, 2021, 6:52 AM • 2 Y

Y by Iora, Coolname

Very nice.
$\left|\dfrac yz-xz\right|\leq 1\Rightarrow \dfrac {y^2}{z^2}+x^2z^2-2xy\leq 1,\left|\dfrac xz+yz\right|\leq 1\Rightarrow \dfrac {x^2}{z^2}+y^2z^2+2xy\leq 1$ Then $\dfrac {y^2}{z^2}+x^2z^2+\dfrac {x^2}{z^2}+y^2z^2\leq 2\Rightarrow (x^2+y^2)(z^2+\dfrac 1{z^2})\leq 2$ By AM-GM $z^2+\dfrac 1{z^2}\ge 2$ . Hence, $x^2+y^2\leq 1$
$x^3+2y\le x^2+2\sqrt{1-x^2}\le x^2+1+1-x^2=2$ Equality Case: $(x,y,z)=(0,1,1)$ or $(x,y,z)=(0,1,-1)$

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#5 h Jun 10, 2021, 8:44 AM • 1 Y

Y by laikhanhhoang_3011

$x^2+y^2\le 1 \to x,y\in [-1,1]$

$x^3+2y\le x^2+y^2+1\le 2$

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sqing

#6 h Jun 10, 2021, 11:51 PM • 2 Y

Y by Mango247, Mango247

BarisKoyuncu wrote:

Let $x,y,z$ be real numbers such that $\left|\dfrac yz-xz\right|\leq 1\text{ and }\left|yz+\dfrac xz\right|\leq 1$ Find the maximum value of the expression $x^3+2y$

Let $x,y>0$ ; $|x-2y| \leqslant \frac{1}{\sqrt{x}}$ and $|y-2x| \leqslant \frac{1}{\sqrt{y}}$ . Find maximum:
$P=x^2+2y$

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#7 h Mar 20, 2023, 6:03 PM • 1 Y

Y by erkosfobiladol

Answer
Proof
Equality

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#8 h Apr 18, 2025, 4:34 PM

Y by

@above can you explain me what ou did in last line?

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Primeniyazidayi

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Primeniyazidayi

#9 h Apr 18, 2025, 5:15 PM

Y by

Namisgood wrote:

@above can you explain me what ou did in last line?

Third inequality gives that $x^2+y^2 \leq 1$ (of course we have $x \leq 1$ .He used this in the first one.It is also trivial that $y^2+1\geq 2y$

This post has been edited 2 times. Last edited by Primeniyazidayi, Apr 18, 2025, 5:16 PM

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