Let $a,b \in R^+$ with $ab = 1$, prove that $$\frac{1}{a^3 + 3b}+\frac{1}{b^3 + 3a}\le \frac12.$$
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Tags: algebra, inequalities
sqing
30.09.2021 16:43
parmenides51 wrote: Let $a,b \in R^+$ with $ab = 1$, prove that $$\frac{1}{a^3 + 3b}+\frac{1}{b^3 + 3a}\le \frac12.$$ By AM-GM, $$\frac{1}{a^3 + 3b}+\frac{1}{b^3 + 3a}=\frac{a}{a^4 + 3}+\frac{b}{b^4+ 3}\le\frac{a}{4a}+\frac{b}{4b} =\frac12.$$
sqing
30.09.2021 16:52
Let $a,b \in R^+$ with $ab = 1$, prove that $$\frac{1}{a^3 + b}+\frac{1}{b^3 + a}\le 1$$$$\frac{1}{a^3 + 2b}+\frac{1}{b^3 + 2a}\le \frac23$$$$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$Where $k\geq 1.$ Let $a,b,c \in R^+$ with $abc = 1$, prove that $$\frac{1}{a^3 +3b}+\frac{1}{b^3 +3c} +\frac{1}{c^3 +3a}\le \frac34 $$
sqing
30.09.2021 17:06
$$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$holds for all positive reals $a,b$ with $ab = 1$, find minimum of $k>0.$
DAVROS
01.10.2021 10:58
sqing wrote: Let $a,b \in R^+$ with $ab = 1$, prove that $\frac{1}{a^3 + b}+\frac{1}{b^3 + a}\le 1$ $LHS = \frac{a^3+b^3+a+b}{a^3b^3+ab+a^4+b^4} = \frac{u(u^2-2)}{(u^2-2)^2} = \frac{u}{u^2-2}$ where $u=a+b$ Inequality is $1-\frac{u}{u^2-2} \ge 0$ or $\frac{(u-2)(u+1)}{u^2-2} \ge 0$, true because $u\ge2\sqrt{ab}=2$
DAVROS
01.10.2021 12:09
sqing wrote: $$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$holds for all positive reals $a,b$ with $ab = 1$, find minimum of $k>0.$
You need $(k+1)(a^3+b^3+k(a+b)) \le 2(a^3b^3+k^2ab+k(a^4+b^4))$ to hold for all $a>0, b>0,ab=1$
Sub $u=a+b, ab=1$ and this is $(k+1)(u^3+(k-3)u) \le 2(1+k^2+k(u^4-4u^2+2))$ for $u \ge 2\sqrt{ab}=2$
Alternatively, $f(v)=v(2kv^3+(15k-1)v^2+(34k-6)v+(-k^2+22k-9)) \ge 0$ where $v=u-2 \ge 0, f(0)=0$
$-k^2+22k-9\ge0 \implies 11-4\sqrt7 \le k \le 11+4\sqrt7$ and for $k$ in this range all terms of are non-negative so $f(v)\ge0$
On the other hand for $k$ outside this range $f$ will decrease initially so $f(v)<0$ for sufficiently small $v$
Inequality is true iff $11-4\sqrt7 \le k \le 11+4\sqrt7$
sqing
01.10.2021 13:16
DAVROS wrote:
sqing wrote:
Let $a,b \in R^+$ with $ab = 1$, prove that $\frac{1}{a^3 + b}+\frac{1}{b^3 + a}\le 1$
$LHS = \frac{a^3+b^3+a+b}{a^3b^3+ab+a^4+b^4} = \frac{u(u^2-2)}{(u^2-2)^2} = \frac{u}{u^2-2}$ where $u=a+b$
Inequality is $1-\frac{u}{u^2-2} \ge 0$ or $\frac{(u-2)(u+1)}{u^2-2} \ge 0$, true because $u\ge2\sqrt{ab}=2$
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sqing
01.10.2021 13:20
DAVROS wrote:
sqing wrote:
$$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$holds for all positive reals $a,b$ with $ab = 1$, find minimum of $k>0.$
answerYou need $(k+1)(a^3+b^3+k(a+b)) \le 2(a^3b^3+k^2ab+k(a^4+b^4))$ to hold for all $a>0, b>0,ab=1$
Sub $u=a+b, ab=1$ and this is $(k+1)(u^3+(k-3)u) \le 2(1+k^2+k(u^4-4u^2+2))$ for $u \ge 2\sqrt{ab}=2$
Alternatively, $f(v)=v(2kv^3+(15k-1)v^2+(34k-6)v+(-k^2+22k-9)) \ge 0$ where $v=u-2 \ge 0, f(0)=0$
$-k^2+22k-9\ge0 \implies 11-4\sqrt7 \le k \le 11+4\sqrt7$ and for $k$ in this range all terms of are non-negative so $f(v)\ge0$
On the other hand for $k$ outside this range $f$ will decrease initially so $f(v)<0$ for sufficiently small $v$
Inequality is true iff $11-4\sqrt7 \le k \le 11+4\sqrt7$
Let $a,b \in R^+$ with $ab = 1.$ Prove that $$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$Where $11-4\sqrt7 \le k \le 11+4\sqrt7.$
sqing
03.10.2021 03:44
sqing wrote:
DAVROS wrote:
sqing wrote:
$$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$holds for all positive reals $a,b$ with $ab = 1$, find minimum of $k>0.$
answerYou need $(k+1)(a^3+b^3+k(a+b)) \le 2(a^3b^3+k^2ab+k(a^4+b^4))$ to hold for all $a>0, b>0,ab=1$
Sub $u=a+b, ab=1$ and this is $(k+1)(u^3+(k-3)u) \le 2(1+k^2+k(u^4-4u^2+2))$ for $u \ge 2\sqrt{ab}=2$
Alternatively, $f(v)=v(2kv^3+(15k-1)v^2+(34k-6)v+(-k^2+22k-9)) \ge 0$ where $v=u-2 \ge 0, f(0)=0$
$-k^2+22k-9\ge0 \implies 11-4\sqrt7 \le k \le 11+4\sqrt7$ and for $k$ in this range all terms of are non-negative so $f(v)\ge0$
On the other hand for $k$ outside this range $f$ will decrease initially so $f(v)<0$ for sufficiently small $v$
Inequality is true iff $11-4\sqrt7 \le k \le 11+4\sqrt7$
Let $a,b \in R^+$ with $ab = 1.$ Prove that $$\frac{1}{a^3 +kb}+\frac{1}{b^3 + ka}\le \frac2{k+1}$$Where $11-4\sqrt7 \le k \le 11+4\sqrt7.$
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