If a, b, c are positive real numbers satisfying 1a+2024+1b+2024+1c+2024=12024 then minimum value of (abc)1/3 is k, where greatest digit in k is________

1a+2024+1b+2024=12024−1c+20241a+2024+1b+2024=c2024(c+2024)Apply A.M. ≥ G.M. in 1a+2024 and 1b+20241a+2024+1b+2024≥2(1(a+2024) . 1(b+2024))1/2⇒c2024(c+2024)≥2(1(a+2024) . 1(b+2024))1/2Similarly, b2024(b+2024)≥2(1(a+2024) . 1(c+2024))1/2 and a2024(a+2024)≥2(1(b+2024) . 1(c+2024))1/2 on multiplying abc20243(a+2024)(b+2024)(c+2024)≥8(a+2024)(b+2024)(c+2024)⇒abc20243≥8⇒(abc)1/3≥4048

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If a, b, c are positive real numbers satisfying $\frac{1}{a + 2024} + \frac{1}{b + 2024} + \frac{1}{c + 2024} = \frac{1}{2024}$ then minimum value of ${(a b c)}^{1 / 3}$ is k, where greatest digit in k is________

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Detailed Solution

$\frac{1}{a + 2024} + \frac{1}{b + 2024} = \frac{1}{2024} - \frac{1}{c + 2024}$

$\frac{1}{a + 2024} + \frac{1}{b + 2024} = \frac{c}{2024 (c + 2024)}$

Apply A.M. $\geq$ G.M. in $\frac{1}{a + 2024}$ and $\frac{1}{b + 2024}$

$\frac{1}{a + 2024} + \frac{1}{b + 2024} \geq 2 {(\frac{1}{(a + 2024)} . \frac{1}{(b + 2024)})}^{1 / 2}$

$\Rightarrow \frac{c}{2024 (c + 2024)} \geq 2 {(\frac{1}{(a + 2024)} . \frac{1}{(b + 2024)})}^{1 / 2}$

Similarly, $\frac{b}{2024 (b + 2024)} \geq 2 {(\frac{1}{(a + 2024)} . \frac{1}{(c + 2024)})}^{1 / 2}$ and $\frac{a}{2024 (a + 2024)} \geq 2 {(\frac{1}{(b + 2024)} . \frac{1}{(c + 2024)})}^{1 / 2}$ on multiplying

$\frac{a b c}{2024^{3} (a + 2024) (b + 2024) (c + 2024)} \geq \frac{8}{(a + 2024) (b + 2024) (c + 2024)}$ $\Rightarrow \frac{a b c}{2024^{3}} \geq 8 \Rightarrow {(a b c)}^{1 / 3} \geq 4048$