$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$

Question

If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?

Answer 1

We have, $$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(x+z) \ge x^3+y^3+z^3+\frac{8}{3}(x+y+z)(xy+yz+xz) = x^3+y^3+z^3+(xy+yz+xz)+(xy+yz+xz)..[8 \text{times}] \ge 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8}$$ Since $$(xy+yz+xz)^2 = x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z) \ge 3xyz(x+y+z)$$ we have, $$ 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8} \ge 9\sqrt[9]{3^8(x^3+y^3+z^3)x^4y^4z^4}$$ And so we have $$ 3 \ge x^4y^4z^4(x^3+y^3+z^3) \Box $$

Answer 2

we can prove this follow $$3^{14}x^4y^4z^4(x^3+y^3+z^3)\le (x+y+z)^{15}$$ then assuming $x+y+z=1$,and denoting $t=3(xy+yz+xz),q=xyz$ $$\Longleftrightarrow 3^{14}q^4(1-t+3q)\le 1\Longleftrightarrow 1-3^{14}q^4(1-t)-3^{15}q^5\ge 0$$ since $$3xyz(x+y+z)\le (xy+yz+xz)^2\Longleftrightarrow q\le\dfrac{t^2}{3^3}$$ so $$1-3^{14}q^4(1-3p)-3^{15}q^5\ge 1-3^{14}\left(\dfrac{t^2}{3^3}\right)^4(1-t)-3^{15}\left(\dfrac{p^2}{3^3}\right)^5=1-9t^8(1-t)-t^{10}$$ and $$1-9t^8+9t^9-t^{10}=(1-t)(t^9+(1+t+t^2+\cdots+t^7-8t^8))=t^9(1-t)+(1-t)^2(1+2t+3t^2+4t^3+\cdots+8t^7)\ge 0$$ since $$t=3(xy+yz+xz)\le(xy+yz+xz)^2=1$$ By done!

Answer 3

Well, I tried and failed to find an answer made of sugar and spice. Here is a frogs and snails answer.

As karafka noted, the AM-GM inequality easily implies that $xyz\leq 1$. So let $\epsilon\geq 0$ be such that $xyz=1-\epsilon$ .

Let us order the variables so that $x\leq y\leq z$. We must have $z\geq 1$ and $x\leq 1$.

Now, it can be checked that \begin{eqnarray*} (xyz)^4(x^3+y^3+z^3) & = & (x+y+z)^3(xyz)^4-3(x+y+z)(xy+xz+yz)(xyz)^4+3(xyz)^5,\end{eqnarray*}

and since $x+y+z=3$ we deduce

\begin{eqnarray*} (xyz)^4(x^3+y^3+z^3) & = & 27(1-\epsilon)^4-9(xy+xz+yz)(1-\epsilon)^4+3(1-\epsilon)^5\\ &=& 3\left(10-\epsilon+3(xy+xz+yz)\right)(1-\epsilon)^4\end{eqnarray*}

Thus we're done if we can show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq \frac{1}{(1-\epsilon)^4}.$

Now, since $$1+\epsilon+\epsilon^2 = \frac{1-\epsilon^3}{1-\epsilon}\leq \frac{1}{1-\epsilon},$$ we can take the fourth power of both sides. Then note that $$1+4\epsilon\leq 1+4\epsilon+10\epsilon^2\leq \frac{1}{(1-\epsilon)^4}.$$

Case 1: $\frac{3}{5}\leq x \leq 1$.

In this case we aim to show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq 1+4\epsilon$, i.e, that $C\geq 0$, where $C=1+4\epsilon - \left(10-\epsilon-3(xy+xz+yz)\right)$.

Since $\frac{3}{5}\leq x\leq 1$ and $z\geq 1$ there exists $a\in[0,\frac{2}{5}]$ such that $x=1-a$ and $b\in[0,2)$ such that $1+b=z$. Hence $y=1+a-b$. Fix such an $a$.

Now, $C$ can be written in terms of $a$ and $b$ as $C=(2-5a)b^2+(5a^2-2a)b+2a^2$. Since $a\leq \frac{2 }{5}$, this corresponds to an upwards parabola in the $C-b$ plane with vertex $\left(\frac{a}{2}, \frac{a^2}{5}(6+5a)\right)$. Hence $C\geq0$ for every choice of $b$.

Case 2: $\frac{3}{25}\leq x\leq \frac{3}{5}$.

In this case, we'll show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq 1+4\epsilon+10\epsilon^2$. For this case, write the right side minus the left in terms of $x,y,z$ to obtain:

$$C=10x^2(yz)^2+(3-25x)(yz)+6+9x-3x^2.$$

Now fix an $x$ in the range. Then $C$ corresponds to upward parabola in "variable" $yz$. Its discriminant is $120x^4-360x^3+385x^2-150x+9.$ Using the quartic formula (...) , this polynomial has real roots $r_1\approx 0.08<\frac{3}{25}$ and $r_2\approx 0.78>\frac{3}{5}$, and is negative in between. Thus for any valid choice of $y,z$, we must have $C>0$.

Case 3: $0<x<\frac{3}{25}$.

The vertex of the parabola from Case $2$ is located at $yz=\frac{25x-3}{20x^2}$. However, this is negative since $x<\frac{3}{25}$. Since the parabola is upwards with $C$-intercept at $C=6+9x-3x^2>6-3>0$, this means that $C>0$ when $yz$ is positive (which it is).

Answer 4

It can be solved by BW method,

I will prove a stronger one: $x^3y^3z^3(x^3+y^3+z^3) \le 3$

$x^3y^3z^3(x^3+y^3+z^3) \le 3 \iff 3^{11}x^3y^3z^3(x^3+y^3+z^3) \le (x+y+z)^{12}$<1>

WOLG, $x=\min[x,y,z], y=x+u,z=x+v, \to u \ge 0, v \ge 0$ put in <1> and rearrange them:

$k_{10}x^{10}+k_{9}x^{9}+k_{8}x^{8}+k_{7}x^{7}+k_{6}x^{6}+k_{5}x^{5}+k_{4}x^{4}+k_{3}x^{3}+k_{2}x^{2}+k_{1}x^{1}+k_0 \ge 0$ <2>

$k_{10}=177147(v^2-uv+u^2) \ge0 \\ k_{9}=39366(v+u)(11v^2-5uv+11u^2) \ge0 \\ k_8=59049(10v^4+13uv^3+33u^2v^2+13u^3v+10u^4) \ge0 \\ k_7=19683(v+u)(34v^4+uv^3+177u^2v^2+u^3v+34u^4) \ge 0 \\ k_6=2187(227v^6+390uv^5+975u^2v^4+2839u^3v^3+975u^4v^2+390u^5v+227u^6) \ge0 \\ k_5=2187(v+u)(88v^6+285uv^5+105u^2v^4+1760u^3v^3+105u^4v^2+285u^5v+88u^6) \ge 0\\ k_4=729(55v^8+440uv^7+811u^2v^6+1622u^3v^5+3850u^4v^4+1622u^5v^3+811u^6v^2+440u^7v+55u^8) \ge0 \\ k_3=27(v+u)(220v^8+1760uv^7+6160u^2v^6+5759u^3v^5+21961u^4v^4+5759u^5v^3+6160u^6v^2+1760u^7v+220u^8) \ge 0 \\ k_2=594(v+u)^{10} \ge0 \\ k_1= 36(v+u)^{11} \ge 0 \\ k_0=(u+v)^{12} \ge 0$

<2> is hold $\implies$ <1> is hold also.

it is trivial that when and only when $u=v=0$ <2> will hold "="

Answer 5

What about this idea: the function is convex on the domain $x,y,z>0$. The constraint $x+y+z=3$ is a convex subdomain. Therefore the function is convex on this subdomain. It can have only one local maximum inside this region, and the symmetry dictates that it must be at $x=y=z=1$.

The maximum of this function can only occur either on the boundary or at the specified point. At the boundary, it equals zero, so the only other candidate for a maximum is $1^41^41^4(1^3+1^3+1^3)=3$

Answer 6

Let's use the arithmetic-geometric inequality.

That gives us $(xyz)^{\frac{1}{4}}≤\frac{1}{4}(x+y+z)\leq\frac{3}{4}$

thus $(xyz)^4\leq(\frac{3}{4})^{16}$

Now all you have to prove is that $x^3+y^3+z^3\leq3*(\frac{4}{3})^{16}$

But $x^3+y^3+z^3\leq 3^3+3^3+3^3=81$ because $x\leq3$ and $y\leq3$ and $z\leq3$.

Thus you have to prove that $3^{15+4}\leq4^{16}$.