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sqing

#1 h Mar 26, 2025, 3:49 AM

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Let $a,b,c\geq 0$ and $a^2+b^2+ ab +24abc\geq\frac{81}{64}$ . Prove that
$a+b+\frac{9}{5}c\geq\frac{9}{8}$ $a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$ $a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$ Let $a,b,c\geq 0$ and $a^2+b^2+ ab +18abc\geq\frac{343}{324}$ . Prove that
$a+b+\frac{6}{5}c\geq\frac{7\sqrt 7}{18}$ $a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$

This post has been edited 1 time. Last edited by sqing, Mar 26, 2025, 4:05 AM

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#2 h Mar 26, 2025, 6:55 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a^2+b^2+ ab +24abc\geq\frac{81}{64}$ . Prove that
$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt[3]{\frac{3}{2}}-\frac{3}{16}$

Since
$(a+b)^2 + ab(24c-1)\ge \frac{81}{64},$ if $24c-1\le 0$ , then $(a+b)^2\ge \frac{81}{64}$ ; hence,
$a+b+\frac{3}{2}c\ge a+b\ge \frac{9}{8} > \frac{9}{8}\sqrt[3]{\frac{3}{2}}-\frac{3}{16}.$ If $24c-1>0$ , then
$ab(24c-1)\le \frac{(a+b)^2(24c-1)}{4},$ so that
$\frac{(a+b)^2(24c+3)}{4}\ge \frac{81}{64}.$ Thus,
$\bigl(8(a+b)\bigr)^2(24c+3)\ge \frac{81}{64}\times 64\times 4=18^2.$ Furthermore, by the AM-GM inequality,
$\bigl(8(a+b)\bigr)^2(24c+3)\le \left(\frac{8(a+b)+8(a+b)+24c+3}{3}\right)^3=\left(\frac{16}{3}\Bigl(a+b+\frac{3}{2}c\Bigr)+1\right)^3.$ Therefore,
$\frac{16}{3}\Bigl(a+b+\frac{3}{2}c\Bigr)+1\ge \sqrt[3]{18^2},$ i.e.,
$a+b+\frac{3}{2}c\ge \frac{9}{8}\sqrt[3]{\frac{3}{2}}-\frac{3}{16}.$

This post has been edited 1 time. Last edited by lbh_qys, Mar 26, 2025, 7:03 AM

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sqing

#3 h Mar 26, 2025, 7:09 AM

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Let $a,b,c\geq 0$ and $a(b+c+ 8bc +1)\geq\frac{28}{27}$ . Prove that
$a+b+c \geq 1$ $50a+57(b+ c) \geq \frac{1400}{27}$ Very very nice.Thank lbh_qys.

sqing wrote:

Let $a,b,c\geq 0$ and $a^2+b^2+ ab +24abc\geq\frac{81}{64}$ . Prove that $a+b+\frac{7}{5}c\geq \frac{9}{8}\sqrt [3]{\frac{7}{5}}-\frac{7}{40}$

Solution of DAVROS:
$ab=0 \implies a+b\ge\frac98 \implies a+b+\frac75c\ge\frac98> \frac{9}{8}\sqrt [3]{\frac{7}{5}}-\frac{7}{40}$ so suppose $ab>0$

Let $u=a+b, v=ab, x=a+b+\frac75c$ then $x \ge u+\frac7{120} + \frac7{120v}\left(\frac{81}{64}-u^2\right)$

$x\ge u+\frac7{120} + \frac7{30u^2}\left(\frac{81}{64}-u^2\right) = u + \frac{189}{640u^2} - \frac7{40} \ge 3\sqrt[3]{\frac{189}{4\times 640}}-\frac7{40}= \frac{9}{8}\sqrt [3]{\frac{7}{5}}-\frac{7}{40}$

$x\ge \frac{9}{8}\sqrt [3]{\frac{7}{5}}-\frac{7}{40}$ with equality at $\frac{u}2= \frac{189}{640u^2}, a=b\implies u=\frac34\sqrt[3]{\frac75}, a=b=\frac{u}2, c=\frac57(x-u)$

This post has been edited 2 times. Last edited by sqing, Mar 27, 2025, 2:55 AM

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sqing

#4 h Mar 26, 2025, 7:59 AM

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Let $a,b,c\geq 0$ and $a^2+b^2+c +ab+10 abc\geq\frac{28}{27}$ . Prove that
$a+b+c \geq 1$ $a+b+\frac{9}{10}c \geq \frac{14}{15}$ $a+b+\frac{11}{10}c \geq \frac{2}{3}\sqrt{\frac{7}{3}}$

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