Find Tuong's minimum

Apply the Cauchy-Schwarz's inequality $\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t} \geq \frac{ (1+1+2+4)^2 }{(x+y+z+t)} = 64$ . If $x=y=\frac{1}{8}$ ; $z=\frac{1}{4}$ and $t=\frac{1}{2}$ , then the equality holds. So the mininum value of $\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}$ is 64.

From the HM-AM (harmonic mean-arithmetic mean) inequality, we can see that

$\begin{aligned} \dfrac{8}{\frac{1}{x}+\frac{1}{y}+2\cdot\frac{1}{z/2}+4\cdot\frac{1}{t/4}} &\leq \dfrac{x+y+2\cdot\frac{z}{2}+4\cdot\frac{t}{4}}{8} \\ &= \frac{x+y+z+t}{8} \\ &= \frac{1}{8} \end{aligned}$

But the LHS of the inequality is just $\frac{8}{\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}}$ , so we have $\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t} \geq 8 \cdot 8 = 64$ . The inequality holds if and only if $x = y = \frac{z}{2} = \frac{t}{4}$ , i.e. when $x = y = \frac{1}{8}, z = \frac{1}{4}, t = \frac{1}{2}$ . So the minimum of $64$ is indeed attainable, and our answer is $64$ .

From the AM-GM inequality, for any positive reals $a, k$ , note $a + \frac{k^2}{a} \ge 2\sqrt{a(k^2/a)} = 2k,$ with equality attained when $a = k$ . Now since $x, y, z, t > 0$ , $\begin{aligned} S &= \frac{1}{x} + \frac{1}{y} + \frac{4}{z} + \frac{16}{t} \\ &= (x+y+z+t)\left(\frac{1}{x} + \frac{1}{y} + \frac{4}{z} + \frac{16}{t}\right) \\ &= (1+1+4+16)+\left(\frac{x}{y}+\frac{y}{x}\right) + \left(\frac{4x}{z} + \frac{z}{x}\right) + \left(\frac{16x}{t} + \frac{t}{x}\right) \\ &\quad + \left(\frac{4y}{z} + \frac{z}{y}\right) + \left(\frac{16y}{t} + \frac{t}{y}\right) + \left(\frac{16z}{t} + \frac{4t}{z}\right) \\ &\ge 22 + 2 + 4 + 8 + 4 + 8 + 4(4) \\ &= 64, \end{aligned}$ with equality attained if $x = y, \quad z = 2x = 2y, \quad t = 4x = 4y, \quad t = 2z.$ With the added constraint that $x + y + z + t = 1$ , we find $x = y = \frac{1}{8}, \quad z = \frac{1}{4}, \quad t = \frac{1}{2}$ satisfies the conditions and achieves the minimum, which is $64$ .

Let the required expression $\frac {1}{x} + \frac {1}{y} + \frac {4}{z} + \frac {16}{t}$ be $P$ .

If we are able to find numbers, such that their sum is $x + y + z + t$ , i.e. 1; and the sum of their reciprocals is $\frac {1}{x} + \frac {1}{y} + \frac {4}{z} + \frac {16}{t}$ , i.e. $P$ ; then using the $A.M. \geq H.M.$ inequality we can find the minimum value of P. Here $A.M.$ is the arithmetic mean of the numbers, and $H.M.$ is the harmonic mean of the numbers.

This means that for each $x, y, z,$ and $t$ , we have to find values of $a$ and $b$ such that $\frac {a}{b}$ is the coefficient of $x, y, z$ and $z$ , and $ab$ is the coefficient for $\frac{1}{x}$ . $a$ is the no. of times term is repeated.

The following equation is obtained for $x$ .

$\frac {a_xx}{b_x} = x$ and $\frac {a_xb_x}{x} = \frac {1}{x}$

$\frac {a_x}{b_x} = 1$ and $a_xb_x = 1$

Similarly, we get the following equations for $y, z,$ and $t$

$\frac {a_y}{b_y} = 1$ and $a_yb_y = 1$

$\frac {a_z}{b_z} = 1$ and $a_zb_z = 4$

$\frac {a_t}{b_t} = 1$ and $a_tb_t = 16$

So, we get the following terms,

$x, y, \frac {z}{2}, \frac {z}{2}, \frac {t}{4}, \frac {t}{4}, \frac {t}{4}, \frac {t}{4}$ .

The $A.M.$ of these numbers is

$\frac {x + y + \frac {z}{2} + \frac {z}{2} + \frac {t}{4} + \frac {t}{4} + \frac {t}{4} + \frac {t}{4} }{8} = \frac{x + y + z + t}{8}$

$= \frac {1}{8}$

The $H.M.$ of these numbers is

$\frac {8}{\frac {1}{x} + \frac {1}{y} + \frac {1}{z} + \frac {1}{z} + \frac {1}{t} + \frac {1}{t} + \frac {1}{t} + \frac {1}{t} + \frac {1}{t}}$

$=\frac {8}{P}$

Using the $A.M. \geq H.M.$ inequality, we get

$\frac {1}{8} \geq \frac {8}{P}$

$P \geq 64$

Thus, we can conclude that the minimum value of $P$ , i.e. $\frac {1}{x} + \frac {1}{y} + \frac {4}{z} + \frac {16}{t}$ is 64 .

By Cauchy Inequality we have (1/x+1/y+4/z+16/t)*(x+y+z+t)>=(sqrt(1)+sqrt(1)+sqrt(4)+sqrt(16))^2=(1+1+2+4)^2=8^2=64

{(1/√x)^2+(1/√y)^2+(2/√z)^2+(4/√t)^2} {(√x)^2+(√y)^2+(√z)^2+(√t)^2}≥{(1/√x)(√x)+(1/√y)(√y)+(2/√z)(√z)+(4/√t)(√t)}^2 (1/x+1/y+4/z+16/t)(1)≥(1+1+2+4)^2 (1/x+1/y+4/z+16/t)≥(8)^2 (1/x+1/y+4/z+16/t)(1)≥64. Done.Sorry i do not know how to use latex XD.

By Cauchy-Schwarz, we have (x+y+z+t)(1/x+1/y+4/z+16/t)>=(1+1+2+4)^2=64. We know that x+y+z+t=1, so we have 1/x+1/y+4/z+16/t>=64.

The minimum must be a local minimum on the hyperplane $x+y+z+t = 1$ in 4-space. It's not on the boundary $xyz = 0$ of the region in question, because the quantity actually increases without bound as it approaches that boundary.

To find the local minima, we use Lagrange multipliers; they say that $(-1/x^2,-1/y^2,-4/z^2,-16/t^2)$ should be parallel to $(1,1,1,1)$ . Since all the variables are positive, we get $x=y, 2x=z, 4x=t$ . This quickly leads to $x=y=1/8, z=1/4, t=1/2.$ Since there is only one extremum in the region, it's either a global minimum or a global maximum, and it's clearly a global minimum because the expression blows up at the boundary. Plugging in, we get an answer of $64$ .

Considere a funcao f(x,y,z,t)=1/x + 1/y + 4/z + 16/t Existe uma restricao: x+y+z+t=1 Derive f(x,y,z,t) com relação a apenas uma das variaveis e iguale a uma constante L, multiplicada a derivada de g(x,y,z,t) com relação apenas a essa mesma variável Para x, temos: A derivada de f apenas em relacao a x é a derivada de 1/x: -1/x² A derivada de g apenas em relação a x é a derivada de x: 1 Logo voce escreve: -1/x² = L 1 Faca o mesmo para todas as variaveis, utilizando o mesmo L: -1/y² = L 1 -4/z² = L 1 -16/t² = L 1 Chame L de -1/c^2 -1/x² = L = -1/c² => x = c -1/y² = L = -1/c² => y= c -4/z² = L = -1/c² => z = 2 c -16/t² = L = -1/c² => t = 4 c Agora, use o fato de x+y+z+t = 1 pra achar c. c+c+2c+4c = 1 8c=1 c=1/8 Assim, temos: x=1/8 y=1/8 z=2 1/8=1/4 t=4 1/8=1/2 f(1/8,1/8,1/4,1/2) = 8+8+4 4+16 2=64

x<=y<z<t and we get x=1/8 y=1/8 z=1/4 t=1/2

1/x + 1/y + 4/z + 16/t = 8 + 8 + 16 + 32 = 64

Once we have $x+y+z+t=1$ and $\frac{1}{x} + \frac{1}{y} + \frac{4}{z} + \frac{16}{t}$ , and x,y,z,t belong to $R^*_+$ , we may conclude that:

I- If one of them belonged to $I$ , the sum would not be 1. So, they belong to $Q$ .

II - Since they're fractions, they may be written as $\frac {a}{b}$ .

III- $t = \frac {a_t}{b_t} \Rightarrow \frac{16}{t} = \frac{16b_t}{a_t}$ . And $b_t > a_t$ .

IV- To reach the lowest sum, $\frac{16b_t}{a_t}$ must be the lowest possible. For this, $b_t$ must be the lowest. So, $b_t = 2 \Rightarrow a_t = 1$ .Then, $t = \frac{1}{2}$ .

V - So $\frac{1}{x} + \frac{1}{y} + \frac{4}{z} + \frac{16}{t} = \frac{1}{x} + \frac{1}{y} + \frac{4}{z} + 32$ .

VI- $x+ y+ z = \frac{1}{2}$ .

Using the same metod, we may infer, firstly, that $z$ would be $\frac{1}{3}$ . But, we have to consider that $b_i$ for $i = {x, y, z, t}$ , must be the lowest possible, and, in this case( $b_z = 3$ ), $b_x = b_y = 12$ .

Obviously, we may say that $x = y$ , because different numbers would make $\frac{b_x}{a_x} + \frac{b_y}{a_y} \neq \frac{2b}{a}$ that would make $a \neq 1$ for every variable( $x,y,z,t$ ).

When we affirm that $b_z > 4$ , $x + y > \frac{1}{4} \Rightarrow x = y > 1/8$ . Then $2b_x + 4b_z +32 > 2 * 8 + 4*4 + 32 = 64$ .

So, the minimum number is 64 .