MHB What is the minimal number satisfying this inequality?

[lfdahl]

Let $a,b$ and $c$ be positive real numbers, and $s = abc$. Find the minimal number, $L$, satisfying: \[ \frac{a^3-s}{2a^3+s}+\frac{b^3-s}{2b^3+s}+\frac{c^3-s}{2c^3+s} \le L \]

 

[Albert1]

Let $a,b$ and $c$ be positive real numbers, and $s = abc$. Find the minimal number, $L$, satisfying: \[ \frac{a^3-s}{2a^3+s}+\frac{b^3-s}{2b^3+s}+\frac{c^3-s}{2c^3+s} \le L \]

my solution:

Spoiler

let :$A=\dfrac{a^3-s}{2a^3+s}+\dfrac{b^3-s}{2b^3+s}+\dfrac{c^3-s}{2c^3+s}$
$=3-(\dfrac{a^3+2s}{2a^3+s}+\dfrac{b^3+2s}{2b^3+s}+\dfrac{c^3+2s}{2c^3+s})$
$\leq 3-3\sqrt [3]{\dfrac{a^3+2s}{2a^3+s}\times\dfrac{b^3+2s}{2b^3+s}\times\dfrac{c^3+2s}{2c^3+s}
}=3-3=0=L$
equality occurs at $a=b=c, s=a^3=b^3=c^3$

 

[lfdahl]

my solution:

Spoiler

let :$A=\dfrac{a^3-s}{2a^3+s}+\dfrac{b^3-s}{2b^3+s}+\dfrac{c^3-s}{2c^3+s}$
$=3-(\dfrac{a^3+2s}{2a^3+s}+\dfrac{b^3+2s}{2b^3+s}+\dfrac{c^3+2s}{2c^3+s})$
$\leq 3-3\sqrt [3]{\dfrac{a^3+2s}{2a^3+s}\times\dfrac{b^3+2s}{2b^3+s}\times\dfrac{c^3+2s}{2c^3+s}
}=3-3=0=L$
equality occurs at $a=b=c, s=a^3=b^3=c^3$

Hi, Albert, and thankyou for your nice solution. Please elaborate on the following identity, which occurs in your answer:

Spoiler

$\sqrt [3]{\dfrac{a^3+2s}{2a^3+s}\times\dfrac{b^3+2s}{2b^3+s}\times\dfrac{c^3+2s}{2c^3+s}
}=1$

 

[Albert1]

Hi, Albert, and thankyou for your nice solution. Please elaborate on the following identity, which occurs in your answer:

Spoiler

$\sqrt [3]{\dfrac{a^3+2s}{2a^3+s}\times\dfrac{b^3+2s}{2b^3+s}\times\dfrac{c^3+2s}{2c^3+s}
}=1$

Spoiler

$\sqrt [3]{\dfrac{a^3+2s}{2a^3+s}\times\dfrac{b^3+2s}{2b^3+s}\times\dfrac{c^3+2s}{2c^3+s}
}=1$
$a=b=c,s=abc=a^3=b^3=c^3$
$\dfrac{a^3+2s}{2a^3+s}=\dfrac{3a^3}{3a^3}=\dfrac{b^3+2s}{2b^3+s}=\dfrac{3b^3}{3b^3}=\dfrac{c^3+2s}{2c^3+s}=\dfrac {3c^3}{3c^3}=1\,\,\, (a,b,c>0)$

 

[lfdahl]

Solution by other:

Spoiler

We prove that $L = 0$. Let

\[ f(a,b,c) = \frac{a^3-s}{2a^3+s}+\frac{b^3-s}{2b^3+s} + \frac{c^3-s}{2c^3+s} \]

Since $f(t,t,t) = 0, L \geq 0$. Let us prove, that $L \leq 0$, equivalently $f(a,b,c) \leq 0$. Since

\[ f(a,b,c) = \frac{-3a^3s^2-3b^3s^2-3c^3s^2+9s^3}{(2a^3+s)(2b^3+s)(2c^3+s)} = \frac{3s^2(3s-a^3-b^3-c^3)}{ (2a^3+s)(2b^3+s)(2c^3+s)} \]

we have to establish the inequality $3s-a^3-b^3-c^3 \leq 0$ ,
which is an arithmetic-geometric inequality for $a^3,b^3$ and $c^3$ . Done.