Gửi bạn bài $5.2$ nha.

Đáp án: 5.1.\(\dfrac 1x+\dfrac 1y\ge \dfrac 4{x+y}\)

Dấu $"="$ xảy ra khi $x=y$

5.2 \(\dfrac 1{2x+y+z}+\dfrac 1{x+2y+z}+\dfrac 1{x+y+2z}\le 1\)

Dấu $"="$ xảy ra khi $x=y=z=\dfrac 34$

Giải thích các bước giải:

\(5.1\)

Vì $x;y$ là các số thực dương. Do đó:

\(\dfrac 1x+\dfrac 1y\ge \dfrac 4{x+y}\)

$↔ \dfrac{x+y}{xy}\ge \dfrac 4{x+y}$

$↔ (x+y)^2\ge 4xy$

$↔x^2+y^2+2xy\ge 4xy$

$↔x^2+y^2-2xy\ge 0$

$↔ (x-y)^2\ge 0$( luôn đúng)

Dấu $"="$ xảy ra khi $x=y$

Vậy \(\dfrac 1x+\dfrac 1y\ge \dfrac 4{x+y}\)

Dấu $"="$ xảy ra khi $x=y$

\(5.2\)

\(\dfrac 1{2x+y+z}=\dfrac 1{(x+y)(x+z)}\le \dfrac 14\left(\dfrac 1{x+y}+\dfrac 1{x+z}\right)\le \dfrac 14\cdot \dfrac 14\left(\dfrac 1x+\dfrac 1y+\dfrac 1x+\dfrac 1z\right)=\dfrac 1{16}\left(\dfrac 2x+\dfrac 1y+\dfrac 1z\right)\qquad(1)\)

CMTT: \(\dfrac 1{x+2y+z}\le \dfrac 1{16}\left(\dfrac 1x+\dfrac 2y+\dfrac 1z\right)\qquad(2)\)

\(\dfrac 1{x+y+2z}\leq\dfrac 1{16}\left(\dfrac 1x+\dfrac 1y+\dfrac 2z\right)\qquad(3)\)

Từ $(1);(2)$ và $(3),$ cộng vế theo vế, ta được

\(\dfrac 1{2x+y+z}+\dfrac 1{x+2y+z}+\dfrac 1{x+y+2z}\le \dfrac 1{16}\left(\dfrac 2x+\dfrac 1y+\dfrac 1z+\dfrac 1x+\dfrac 2y+\dfrac 1z+\dfrac 1x+\dfrac 1y+\dfrac 2z\right)\)

\(\to \dfrac 1{2x+y+z}+\dfrac 1{x+2y+z}+\dfrac 1{x+y+2z}\le \dfrac 1{16}\cdot 4\cdot \left(\dfrac 1x+\dfrac 1y+\dfrac 1z\right)=\dfrac 1{16}\cdot 4\cdot 4=1\)

Dấu $"="$ xảy ra khi $x=y=z=\dfrac 34$

Vậy \(\dfrac 1{2x+y+z}+\dfrac 1{x+2y+z}+\dfrac 1{x+y+2z}\le 1\)

Dấu $"="$ xảy ra khi $x=y=z=\dfrac 34$