Prove this $4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$ 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
$$a^3+b^3+c^3+d^3\ge 4$$
 A: 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 $a=b=c=d=1$

A: 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:
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\leq 20$$
