v_Enhance wrote: Prove that for positive reals $a$, $b$, $c$ we have \[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \]

\[ 3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \]

v_Enhance wrote: Prove that for positive reals $a$, $b$, $c$ we have \[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] Prove that for positive reals $a$, $b$ we have \[ 2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}. \] Prove that for positive reals $a$, $b$, $c$, $d$ we have \[ 4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}. \]

sqing wrote: v_Enhance wrote: Prove that for positive reals $a$, $b$, $c$ we have \[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] Prove that for positive reals $a$, $b$ we have \[ 2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}. \] Prove that for positive reals $a$, $b$, $c$, $d$ we have \[ 4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}. \] hello, your inequality is equivalent to $\frac{1}{4}(a-b)^2(49a^2-2ab+49b^2)\geq 0$ after two times squaring. Sonnhard.

Dr Sonnhard Graubner wrote: sqing wrote: v_Enhance wrote: Prove that for positive reals $a$, $b$, $c$ we have \[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] Prove that for positive reals $a$, $b$ we have \[ 2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}. \] Prove that for positive reals $a$, $b$, $c$, $d$ we have \[ 4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}. \] hello, your inequality is equivalent to $\frac{1}{4}(a-b)^2(49a^2-2ab+49b^2)\geq 0$ after two times squaring. Sonnhard. $ 2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}\iff \frac{1}{4}(a-b)^2(49a^2-2ab+49b^2)\geq 0.$ Thanks.

sqing wrote: \[ 3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] $ 3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\iff a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge 6abc.$ $\frac{8\sqrt[3]{abc} + \sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}=\frac{\sqrt[3]{abc} +\sqrt[3]{abc} +\cdots+\sqrt[3]{abc} +\sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}$ $\le \sqrt[3]{\frac{abc+abc+\cdots+abc+\frac{a^ 3+b^3+c^3}{3}}{9}}= \sqrt[3]{\frac{8abc+\frac{a^ 3+b^3+c^3}{3}}{9}}$ $\Rightarrow 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

sqing wrote: sqing wrote: \[ 3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] $ 3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\iff a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge 6abc.$ $\frac{8\sqrt[3]{abc} + \sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}=\frac{\sqrt[3]{abc} +\sqrt[3]{abc} +\cdots+\sqrt[3]{abc} +\sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}$ $\le \sqrt[3]{\frac{abc+abc+\cdots+abc+\frac{a^ 3+b^3+c^3}{3}}{9}}= \sqrt[3]{\frac{8abc+\frac{a^ 3+b^3+c^3}{3}}{9}}$ $\Rightarrow 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$ $9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$ By Holder, ${\underbrace{(1+1+\cdots +1)}_{9}}^{2/3}\left(8abc+\frac{a^3+b^3+c^3}{3}\right)^{1/3}\ge 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}. $ $\Rightarrow 9^{2/3}.9^{1/3}\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}. $ So, it is done

v_Enhance wrote: Prove that for positive reals $a$, $b$, $c$ we have \[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=228831

My solution

This inequality is <=> with :

sqing wrote: Prove that for positive reals $a$, $b$, $c$, $d$ we have \[ 4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}. \] is it true?

v_Enhance wrote: Prove that for positive reals $a$, $b$, $c$ we have \[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \] Stronger inequality: $3\left( {a + b + c} \right) \ge 8\sqrt[3]{{\frac{{\left( {a + b + c} \right)\left( {ab + bc + ca} \right)}}{9}}} + \sqrt[3]{{\frac{{{a^3} + {b^3} + {c^3}}}{3}}}$.

thanks you!

sqing wrote: Prove that for positive reals $a$, $b$, $c$, $d$ we have \[ 4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}. \] Is it true. if it is true then how to prove it?

By $uvw$ we can show that the following inequality is also true. Let $a$, $b$, $c$ and $d$ be non-negative numbers. Prove that: $$2(a+b+c+d)\geq7\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}+\sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$$

Let $4u=a+b+c+d$, $6v^2=ab+cd+ac+bd+ad+bc$, $4w^3=abc+bcd+cda+dab$ and $t^4=abcd$, then we need to prove $$8u\geq 7w+\sqrt[4]{64u^4-96u^2v^2+18v^4+16uw^3-t^4}$$or $$8u-7w\geq \sqrt[4]{64u^4-96u^2v^2+18v^4+16uw^3-t^4}$$or $$(8u-7w)^4\geq 64u^4-96u^2v^2+18v^4+16uw^3-t^4$$or $4032u^4+96u^2v^2-18v^4-14336u^3w+18816u^2w^2-10992uw^3+2401w^4+t^4$. Since $t^4\geq4uw^3-3v^4$. thus $4032u^4+96u^2v^2-18v^4-14336u^3w+18816u^2w^2-10992uw^3+2401w^4+t^4\geq 4032u^4+96u^2v^2-21v^4-14336u^3w+18816u^2w^2-10988uw^3+2401w^4$. By Rolle's theorem $(X-a)(X-b)(X-c)(X-d)'=4(X^3-3uX^2+3v^2X-w^3)$ has three real roots. Let $3u=x+y+z$, $3v^2=xy+yz+zx$ and $w^3=xyz$, then we need to prove $f(v^2)\geq 0$, where $f(v^2)$ is concave, which means enough to check when two of variables are equal. WLOG $y=z=1$. then it's equivalent to $(x - 1)^2 (1344 x^{10} + 2688 x^9 - 10304 x^8 - 12352 x^7 + 42048 x^6 + 10432 x^5 - 87208 x^4 + 40944 x^3 + 61891 x^2 - 71038 x + 21825)\geq 0$, which is true. P.S- in the last part I substituted $x\rightarrow x^3$.

WLOG let $abc=1$ since this equation is homogeneous. We note that weighted power mean when we set weights $w_1=\frac19$ and $w_2=\frac89$ gets that letting the positive real numbers $a_1=\frac13(a^3+b^3+c^3)$ and $a_2=1$ gets when we have the power-mean number as $1$ vs $\frac13$, we have the inequality $$\frac19 \cdot \frac13(a^3+b^3+c^3) + \frac89 \ge \left[\frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac89\right]^3$$which means since $$\frac19 \cdot \frac13(a^3+b^3+c^3) = \frac{a^3+b^3+c^3+24abc}{27}\le \frac{\sum_{\text{cyc}}{a^3}+3\sum_{\text{sym}}{a^2b}+6abc}{27} = \frac{1}{27}(a+b+c)^3$$by Muirhead's Inequality over $3\sum_{\text{sym}}{a^2b} \ge 3 \cdot \sum_{\text{sym}}{abc} = 18abc$. As a result, $$\frac{(a+b+c)^3}{27} \ge \left[\frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac89\right]^3$$so we can take the cube root of both sides getting $$\frac13(a+b+c) \ge \frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac89=\frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac{8abc}9$$and multiplying both sides by $9$ gets $$3(a+b+c) \ge \sqrt[3]{\frac13(a^3+b^3+c^3)} + 8abc$$which is our desired inequality.

By Power Mean, we see that \begin{align*} \frac{8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}}}{9} & \le \sqrt[3]{\frac{8abc + \frac{a^3 + b^3 + c^3}{3}}{9}} \end{align*} and so \begin{align*} 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} & \le 9 \cdot \sqrt[3]{\frac{8abc + \frac{a^3 + b^3 + c^3}{3}}{9}}\\ & = 3\sqrt[3]{a^3 + b^3 + c^3 + 24abc}\\ & \le 3\sqrt[3]{(a+b+c)^3}\\ & = 3(a+b+c) \end{align*} as desired. $\square$

By weighted power mean with weights $\tfrac19, \tfrac89$ and reals $$\frac{a^3+b^3+c^3}{3}, abc$$we have \begin{align*} \frac{\frac{a^3+b^3+c^3}{3}+8abc}{9}&\ge \left(\frac{\sqrt[3]{\frac{a^3+b^3+c^3}{3}} + 8\sqrt[3]{abc}}{9}\right)^3.\\ \end{align*}After doing some manipulations, we see it suffices to show $$a^3+b^3+c^3+24abc \le (a+b+c)^3$$or $$18abc \le 3a^2b+3a^2c+3ab^2+3ac^2+3bc^2+3b^2c.$$Since $(2,1,0)\succ(1,1,1)$ by Muirhead, we have $$\sum_{sym}a^2b \ge \sum_{sym}abc=6abc,$$so multiplying be $3$, we are done. $\blacksquare$

Set $f(x)=\sqrt[3]{x}$ which is clearly concave. By Jensen's, $$9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$so it suffices so show the stronger \begin{align*}3(a+b+c)&\ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\\ \iff \frac{(a+b+c)^3}{27}&\ge \frac{\frac{3\cdot 8abc}{3}+\frac{a^3+b^3+c^3}{3}}{9}\\ \iff (a+b+c)^3&\ge a^3+b^3+c^3+24abc\\ \iff a^3+b^3+c^3+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)+6abc &\ge a^3+b^3+c^3+24abc\\ \iff a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\ge 6abc \end{align*}which is obvious by Muirhead/AM-GM.

Weighted power mean with weights $\frac{8}{9}$ and $\frac{1}{9}$, and 2 numbers $\frac{a^3+b^3+c^3}{3},abc$ yields $\frac{1}{9}\left(\frac{a^3+b^3+c^3}{3}\right)+\frac{8}{9}(abc) \geq \frac{1}{9}\sqrt{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt{abc}$. Thus, we wish to prove $a^3+b^3+c^3+24abc \le (a+b+c)^3$, or $a^2b+a^2c+ab^2+ac^2+bc^2+b^2c \ge 6abc$. This follows directly from Muirhead, so we are done.

sqing wrote: Prove that for positive reals $a$, $b$ we have \[ 2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}. \] Good.

Second is much harder with this method.

Setting $s=a+b+c, q=\sqrt{3ab+3bc+3ca}, p=3\sqrt[3]{abc}$ (the idea is to use the three symmetric expressions for $a,b,c$, so that $s=p=q$ if $a=b=c$ and then get rid of the cube root, using $s,q,p$). We can transfrom the inequality as follows: $$3(a+b+c)=3s,\quad 8\sqrt[3]{abc}=\frac{8}{3}p,\quad a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ca)(a+b+c)+3abc=s^3-q^2s+\frac{1}{9}p^3$$$$\implies 3(a+b+c)\geq 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\iff 3s-\frac{8}{3}p\geq \sqrt[3]{\frac{s^3-q^2s+\frac{1}{9}p^3}{3}}$$$$\iff 9s-8p\geq \sqrt[3]{9s^3-9q^2s+p^3}\iff 729s^3-1944s^2p+1728sp^2-512p^3\geq 9s^3-9q^2s+p^3$$$$\iff 720s^3+9q^2s+1728sp^2-1944s^2p-513p^3\geq 0$$However, by AM-GM we have that $s\geq q\geq p$ and we can finish the solution substituting $q$ with $p$ and factoring the polynomial as we know that $s=p$ will be a root as obviously the inequality case is achieved if $a=b=c\implies s=p$: $$720s^3+9q^2s+1728sp^2-1944s^2p-513p^3\geq 720s^3+9p^2s+1728sp^2-1944s^2p-513p^3=9(s-p)(4s-3p)(20s-19p)\geq 0$$The last inequality follows from $s\geq p>0\implies 4s-3p>0, 20s-19p>0, s-p\geq 0$, thus we've solved the problem! (also we know that the inequality reaches equality iff $s=p\implies a=b=c$)

Consider the following set of $9$ elements: $$ \left\{8\sqrt[3]{abc},8\sqrt[3]{abc},\dots,8\sqrt[3]{abc}, \thickspace 8\cdot \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right\} \, , $$Applying the inequality between the cubic and arithmetic means to the elements of the set, one gets $$ \frac{64\sqrt[3]{abc}+8\cdot \sqrt[3]{\frac{a^3+b^3+c^3}{3}}}{9}\leq \sqrt[3]{\frac{8\cdot (8\sqrt[3]{abc})^3+\left(8\cdot \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right)^3}{9}}, $$which is equivalent to $$ 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\leq 3\cdot \sqrt[3]{24 \, abc+a^3+b^3+c^3} $$ Hence in order to prove the heading it suffices to show $$ 3(a+b+c)\geq 3\cdot \sqrt[3]{24 \, abc+a^3+b^3+c^3} $$And this is clear, because $$ (a+b+c)^3\geq a^3+b^3+c^3+24abc \iff 3\cdot (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)\geq 18abc \iff $$$$ \iff \frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\geq 6, $$which is true by AM-GM.

Can we generalize this inequality into $n$-variables? Let $n\geq 2$ be an integer and $a_1,a_2,\cdots,a_n$ be positive reals. Find min $\lambda$ such that $$\sum_{i=1}^n a_i \geq \lambda \sqrt[n]{\prod_{i=1}^n a_i} + (1 - \lambda) \sqrt[n]{\frac{1}{n}\sum_{i=1}^n a_i^n}$$

Notice that since $f(x)=\sqrt[3]{x}$ is concave so by Jensen's we have \[ \frac{8f(abc)+f\left(\frac{a^3+b^3+c^3}{3}\right)}{9}\le f\left(\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}\right). \]Then we have \begin{align*} 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}} &\le 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}} \\ &= 3\sqrt[3]{a^3+b^3+c^3+24abc} \\ &\le 3\sqrt[3]{(a+b+c)^3} \\ &= 3(a+b+c). \end{align*}

$$(a+b+c)^3\geq a^3+b^3+c^3+24abc$$$$a^3+b^3+c^3=3x^3$$$$abc=y^3$$$$81(x^3+8y^3)\geq (x+8y)^3$$Remaining easy.

Taco12 wrote: Weighted power mean with weights $\frac{8}{9}$ and $\frac{1}{9}$, and 2 numbers $\frac{a^3+b^3+c^3}{3},abc$ yields $\frac{1}{9}\left(\frac{a^3+b^3+c^3}{3}\right)+\frac{8}{9}(abc) \geq \frac{1}{9}\sqrt{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt{abc}$. Thus, we wish to prove $a^3+b^3+c^3+24abc \le (a+b+c)^3$, or $a^2b+a^2c+ab^2+ac^2+bc^2+b^2c \ge 6abc$. This follows directly from Muirhead, so we are done. interesting, but I don't understand why the last inequality ($a^3+b^3+c^3+24abc \le (a+b+c)^3$)is what is missing to solve the problem. anyone could explain me please?

By Power Mean with weights $\frac 89$ and $\frac 19$, $$\left(\frac 89\sqrt[3]{abc} + \frac 19\sqrt[3]{\frac{a^3+b^3+c^3}3}\right)^3 \leq \frac 89 abc + \frac 19 \cdot \frac{a^3+b^3+c^3}3.$$Thus, it suffices to show that $$\left(\frac{a+b+c}3\right)^3 \geq \frac 19\left(8abc+\frac{a^3+b^3+c^3}3\right) \iff (a+b+c)^3 \geq 24abc+a^3+b^3+c^3.$$This is evident.

Note how most of these solutions were just like Evan’s in OTIS Excerpts Assim584 wrote: Taco12 wrote: Weighted power mean with weights $\frac{8}{9}$ and $\frac{1}{9}$, and 2 numbers $\frac{a^3+b^3+c^3}{3},abc$ yields $\frac{1}{9}\left(\frac{a^3+b^3+c^3}{3}\right)+\frac{8}{9}(abc) \geq \frac{1}{9}\sqrt{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt{abc}$. Thus, we wish to prove $a^3+b^3+c^3+24abc \le (a+b+c)^3$, or $a^2b+a^2c+ab^2+ac^2+bc^2+b^2c \ge 6abc$. This follows directly from Muirhead, so we are done. interesting, but I don't understand why the last inequality ($a^3+b^3+c^3+24abc \le (a+b+c)^3$)is what is missing to solve the problem. anyone could explain me please? This is a very well known result, and easy to prove. Expanding, it suffices to prove that $3(a^2b+ab^2+b^2c+c^2b+a^2c+c^2a)\geq 3(6abc)$, which follows immediately from dividing by 3 and AM-GM over all of the terms. Muirhead with (2,1,0) majorizing (1,1,1) would also suffice.

This was rough. Power mean inequality gives, \[ \left( \frac{8}{9} \cdot \sqrt[3]{abc} + \frac{1}{9}\cdot \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} \right)^3 \leq \frac{8}{9} \cdot abc + \frac{1}{9} \cdot \frac{a^3+b^3+c^3}{3}. \]Then we wish to show, \begin{align*} (a+b+c)^3 &\geq 24abc + a^3+b^3+c^3\\ 3\sum_{sym} a^2b &\geq 18abc \end{align*}which is just Muirheads.

By the Power Mean, we have that \[\frac{\frac{a^3+b^3+c^3}{3}+8abc}{9}\geq\left(\frac{\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+8\sqrt[3]{abc}}{9}\right)^3.\]Which is equivalent to \[81(\frac{a^3+b^3+c^3}{3}+8abc)\geq (\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+8\sqrt[3]{abc})^3,\]meaning that we just have to prove that \[27(a+b+c)^3\geq 81(\frac{a^3+b^3+c^3}{3}+8abc),\]or \[(a+b+c)^3 \geq a^3+b^3+c^3+24abc.\]Expanding, this is equivalent to proving that \[a^2b+b^2a+a^2c+c^2a+b^2c+c^2b\geq 6abc,\]which is true by Muirhead's, since $(2,1,0)$ majorizes $(1,1,1)$, finishing the problem.

Nice problem. first we take $f(x) = x^{\frac{1}{3}} , x>0 $ Note that $f'= 1/3 x^{-2/3} >0 $ and $ f" = (1/3)\cdot(-2/3) x^{-5/3} < 0 $ and so $f$ is concave on $(0,\infty)$ Dividing both sides by $9$ the inequality is reduced to $$ \frac{a+b+c}{3} \geq \frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt[3]{abc} $$Now note that by Jensen's inequality, $$\frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt[3]{abc}$$$$ = \frac{8}{9} f (abc)+\frac{1}{9} f (\frac{a^3+b^3+c^3}{3}) $$ $$\leq f(\frac{a^3+b^3+c^3+24abc}{27}) $$Now note that $a^3+b^3+c^3+24abc \leq (a+b+c)^3$ (do AM GM , Muirhead whatever) and $f$ is increasing so $$ f(\frac{a^3+b^3+c^3+24abc}{27}) \leq f(\frac{(a+b+c)^3}{27}) = \frac{a+b+c}{3}$$$$\blacksquare$$

Powerful problem, By Weighted Power Mean with weights $\tfrac{1}{9}$, and $\tfrac{8}{9}$ $$\frac{8abc}{9}+\frac{a^{3}+b^{3}+c^{3}}{27}\geq \left(\frac{8}{9} \sqrt[3]{abc}+ \frac{1}{9} \sqrt[3]{\tfrac{a^3+b^3+c^3}{3}} \right)^{3}$$$$3 \sqrt[3]{a^{3}+b^{3}+c^{3}+24abc} \geq 8\sqrt[3]{abc} + \sqrt[3]{\tfrac{a^3+b^3+c^3}{3}}$$So, it suffices to show: $$3(a+b+c) \geq 3 \sqrt[3]{a^{3}+b^{3}+c^{3}+24abc}$$$$(a+b+c)^{3} \geq a^{3}+b^{3}+c^{3}+24abc$$Which is obvious by expansion and muirhead $\square$

By wieghted power mean we get \[\left(\frac{8\sqrt[3]{abc}}{9} + \frac{1}{9} \cdot \frac{\sqrt[3]{a^3+b^3+c^3}}{3}\right)^3 \leq \frac{a^3+b^3+c^3}{27}+\frac{8abc}{9}\]Thus it remains to show that \[\frac{a^3+b^3+c^3}{27}+\frac{8abc}{9} \leq \frac{(a+b+c)^3}{27} \Rightarrow a^3+b^3+c^3 +24 abc \leq (a+b+c)^3\]This is obvious from Muirhead.

Apply weighted power mean on $P(1) \geq P(\frac13)$ with weights $\frac89$ and $\frac19$ and positive reals $abc$ and $\frac{a^3+b^3+c^3}{3}$. After dividing both sides by $9$, the RHS is then $\sqrt[3]{P(\frac13)}$ so it suffices to prove $$\frac{a+b+c}{3} \geq \sqrt[3]{P(\frac13)}$$or $$(\frac{a+b+c}{3})^3 \geq P(1)$$by our weighted power mean. Now $P(1) = \frac{24abc+a^3+b^3+c^3}{27}$ so we just need to prove that $$(a+b+c)^3 \geq 24abc + a^3+b^3+c^3$$$$3 \sum_{\text{sym}} ab^2 \geq 18abc$$after expansion which is true by a simple AMGM.