let $a,b,c,d\in R$,and such $a^2+b^2+c^2+d^2=4$, prove or disprove $$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$$
I try use Cauchy-Schwarz inequality have $$4(a+b+c+d)\le 4\sqrt{4(a^2+b^2+c^2+d^2)}=16$$ but $$(a^3+b^3+c^3+d^3)^2(1+1+1+1)\ge (a^2+b^2+c^2+d^2)^3$$ so we have $$a^3+b^3+c^3+d^3\ge 4$$
Clearly, we only need to prove the case when $a, b, c, d \ge 0$.
After homogenization, it suffices to prove that, for all $a, b, c, d\ge 0$, $$(a^2 + b^2 + c^2 + d^2)(a + b + c + d) + a^3 + b^3 + c^3 + d^3 \le \frac52 (a^2 + b^2 + c^2 + d^2)^{3/2}.$$
WLOG, assume that $d = \max(a, b, c, d)$.
If $d = 0$, the inequality is true.
In the following, assume that $d > 0$. WLOG, assume that $d = 1$. It suffices to prove that, for all $a, b, c \in [0, 1]$, $$(a^2 + b^2 + c^2 + 1)(a + b + c + 1) + a^3 + b^3 + c^3 + 1 \le \frac52 (a^2 + b^2 + c^2 + 1)^{3/2}.$$
Let $p = a + b + c$, $q = a^2 + b^2 + c^2$ and $r = abc$. We have $$0 \le q \le 3, \quad \sqrt{q} \le p \le \sqrt{3q}. \tag{1}$$
Using $a^3 + b^3 + c^3 = -\frac12 p^3 + \frac32 pq + 3r$, the inequality is written as $$(q + 1)(p + 1) -\frac12 p^3 + \frac32 pq + 3r + 1 \le \frac52(q + 1)^{3/2}. \tag{2}$$
Using $(ab + bc + ca)^2 \ge 3(a + b + c)abc$ (easy) which is written as $(\frac{p^2 - q}{2})^2 \ge 3pr$, we have $$r \le \frac{(p^2 - q)^2}{12p}. \tag{3}$$
Using (3), it suffices to prove that $$(q + 1)(p + 1) -\frac12 p^3 + \frac32 pq + 3\cdot \frac{(p^2 - q)^2}{12p} + 1 \le \frac52(q + 1)^{3/2}$$ or $$2pq + p + q + 2 - \frac14p^3 + \frac{q^2}{4p} \le \frac52(q + 1)^{3/2}. \tag{4}$$
Let $f(p) = 2pq + p + q + 2 - \frac14p^3 + \frac{q^2}{4p}$. We have $$f'(p) = 2q + 1 - \frac34 p^2 - \frac{q^2}{4p^2}.$$ We have, for all $\sqrt{q} \le p \le \sqrt{3q}$, $$f''(p) = - \frac{3p^4 - q^2}{2p^3} \le 0.$$ Also, $f'(\sqrt{3q}) = 1 - q/3 \ge 0.$ Thus, $f'(p) \ge 0$ for all $\sqrt{q} \le p \le \sqrt{3q}$.
It suffices to prove that, for all $0 \le q \le 3$, $$f(\sqrt{3q}) \le \frac52(q + 1)^{3/2}$$ or $$\frac{4}{\sqrt 3} q^{3/2} + \sqrt{3q} + q + 2 \le \frac52(q + 1)^{3/2}$$ which is true (not difficult).
We are done.
The inequality can be generalized to $n$ variables.
Problem 1: Let $n\ge 4$. Let $a_i \ge 0, i=1, 2, \cdots, n$ with $\sum_{i=1}^n a_i^2 = n$. Prove that $$n\sum_{i=1}^n a_i + \sum_{i=1}^n a_i^3 \le n^2 + n.$$
Using Vasc's Equal Variable Theorem (Corollary 1.9, [1]), it suffices to prove the case when $a_1 = a_2 = \cdots = a_{n - 1}$.
Let $a_1 = a_2 = \cdots = a_{n - 1} = x \ge 0$ and $a_n = y \ge 0$. We have $(n - 1)x^2 + y^2 = n$ which results in $0 \le x \le \sqrt{n/(n - 1)}$ and $$y = \sqrt{n - (n - 1)x^2}.$$
It suffices to prove that $$n [(n - 1)x + y] + (n - 1)x^3 + y^3 \le n^2 + n$$ or $$n^2 + n - (n - 1)x^3 - (n^2 - n)x \ge (2n + x^2 - nx^2)\sqrt{n - (n - 1)x^2}.$$ It is easy to prove that $\mathrm{LHS} \ge 0$. It suffices to prove that $$[n^2 + n - (n - 1)x^3 - (n^2 - n)x]^2 \ge (2n + x^2 - nx^2)^2[n - (n - 1)x^2]$$ or $$n(x - 1)^2(n - 1)[(n - 1)(x^4 + 2x^3 + n) - 4nx] \ge 0$$ which is true (not difficult).
Reference:
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007. Online: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
We can solve it using the Lagrangian mutliplier:
Given that: $a^2 + b^2 + c^2 + d^2 = 4$ and $a,b,c,d \in \mathbb{R}.$ Prove $$\displaystyle 4(a+b+c+d) + (a^3 + b^3 + c^3 + d^3 ) \le 20 $$
Let us define the constraint function: $\displaystyle g(x) = a^2 + b^2 + c^2 + d^2.$
And If we define a set $\displaystyle \bar{U} =\{ (a,b,c,d) : a^2 + b^2 + c^2 + d^2 \le 1000\}$
Then our constraint set will be $$\displaystyle \bar{S} =\{ x \in \bar{U} : g(x) = 4 \} $$
Also our objective function is : $\displaystyle f(x) = 4(a+b+c+d) + (a^3 + b^3 + c^3 + d^3 ) $.
What we are left to do is, maximize $f(x).$
Now we define a Lagrangian function as follows:
$$\displaystyle \mathcal{L}(x,\lambda) = f(x) - \lambda g(x)$$
Thus to find maxima of the Lagrangian function, we need to find $(x, \lambda) $ that it satify : $\displaystyle \nabla \mathcal{L}(x,\lambda) = 0 \ \text{and} \ \nabla g(x) \ne 0 $.
Futher we get :
$$ \displaystyle \left( \begin{array}[c] *3a^2 + 4 \\ 3b^2 + 4 \\ 3c^2 + 4 \\ 3d^2 + 4 \\ \end{array} \right) = \lambda \left( \begin{array}[c] *2a \\ 2b \\ 2c \\ 2d \\ \end{array} \right) $$
Hence $\displaystyle (a,b,c,d) \in \{ \frac{\lambda - \sqrt{\lambda ^2 -12} }{3} , \frac{\lambda + \sqrt{\lambda ^2 -12} }{3} \} $
Therefore we have deal with four cases and the constraint, with the condition that $\displaystyle \lambda \in \mathbb{R} \geq \sqrt{12} $:
$\displaystyle 1. \ a=b=c=d \\ 2. \ a=b\ne c=d \\ 3. \ a=b=c \ne d$
On solving for $\lambda$ for each of the cases we get $ \lambda $ for case $2$ and $3$ dosen't satisfy the above metioned condition.
Hence the only solutions we get is for case $1$ : $ a = \pm 1 \ \text{and} \ \lambda = 3.5$ . Also at this point $\displaystyle \nabla g(x) \ne 0 $
Finally we have $\displaystyle \mathcal{L}(1,1,1,1,3.5) , \mathcal{L}(-1,-1,-1,-1,-3.5) $ as stationary points and fuction $f(x)$ will attain maxima at $(1,1,1,1)$.
Given that: $a^2 + b^2 + c^2 + d^2 = 4$ and $a,b,c,d \in \mathbb{R} \implies$ $$\displaystyle 4(a+b+c+d) + (a^3 + b^3 + c^3 + d^3 ) \le 20 $$ with equality at $a=b=c=d=1$
You found $a^3+b^3+c^3+d^3\geq 4$
Now using Tchebychev inequality we have:
$$(a+b+c+d)(a^2+b^2+c^2+d^2)\leq 4(a^3+b^3+c^3+d^3)$$
Or:
$$4(a+b+c+d)\leq 4(a^3+b^3+c^3+d^3)$$
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\leq 5(a^3+b^3+c^3+d^3)$$
But:
$$5(a^3+b^3+c^3+d^3)\geq 20$$
therefore:
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\leq 20$$
