lỗi đề kìa bạn

Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh

\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Áp dụng holder ta có:

\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)

\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)

\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)

Từ đây ta cần chứng minh:

\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )

Vậy có ĐPCM

bài hơi khoai

Không mất tính tổng quát giả sử \(c=max\left\{a,b,c\right\}\)

\(\Rightarrow2c\ge a+b\)

\(\Rightarrow c\ge\frac{a+b}{2}\)

Từ giả thiết \(\Rightarrow a,b\le1\)

\(\Rightarrow ab\le1\)( *)

Đặt \(P=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-\frac{5}{2}\)

\(=\frac{1}{a+b}+\frac{1}{b+\frac{1-ab}{a+b}}+\frac{1}{a+\frac{1-ab}{a+b}}-\frac{5}{2}\)

Đặt \(S=\frac{1}{a+b+\frac{1}{a+b}}+a+b+\frac{1}{a+b}-\frac{5}{2}\)

Xét hiệu \(P-S=\)\(\frac{1}{a+b}+\frac{1}{b+\frac{1-ab}{a+b}}+\frac{1}{a+\frac{1-ab}{a+b}}-\frac{5}{2}-\)\(-\frac{1}{a+b+\frac{1}{a+b}}-a-b-\frac{1}{a+b}+\frac{5}{2}\)

\(=\frac{1}{\frac{ab+b^2+1-ab}{a+b}}+\frac{1}{\frac{a^2+ab+1-ab}{a+b}}-\frac{1}{\frac{\left(a+\right)^2+1}{a+b}}-\left(a+b\right)\)

\(=\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}-\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\)

Ta sẽ chứng minh \(\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}-\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\ge0\)

\(\Leftrightarrow\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}\ge\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\)

\(\Leftrightarrow\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge1+\frac{1}{1+\left(a+b\right)^2}\)

\(\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2+\left(a+b\right)^2}{1+\left(a+b\right)^2}\)

\(\Rightarrow\left(2+b^2+a^2\right)\left[1+\left(a+b\right)^2\right]\ge\left[2+\left(a+b\right)^2\right]\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow2+2\left(a+b\right)^2+\left(a+b\right)^2\left(a^2+b^2\right)+a^2+b^2\ge\left[2+\left(a+b\right)^2\right]\left(1+a^2+b^2+a^2b^2\right)\)

\(\Leftrightarrow2+2\left(a+b\right)^2+\left(a+b\right)^2\left(a^2+b^2\right)+a^2+b^2-2a^2b^2-\left(a+b\right)^2\left(a^2+b^2\right)-\left(a+b\right)^2a^2b^2\)\(-2-2\left(a^2+b^2\right)-\left(a+b^2\right)\ge0\)

\(\Leftrightarrow-2a^2b^2-\left(a+b\right)^2a^2b^2+a^2+b^2-\left(a+b\right)^2\ge0\)

\(\Leftrightarrow ab\left[ab\left(a+b\right)^2+2ab-2\right]\le0\)

\(\Leftrightarrow ab\left(a+b\right)^2+2ab-2\le0\)( do a,b \(\ge0\))

\(\Leftrightarrow ab\left(a+b\right)^2\le2\left(1-ab\right)\)

\(\Leftrightarrow ab\left(a+b\right)^2\le2c\left(a+b\right)\) (1)

Mà \(c\ge\frac{a+b}{2}\)

\(\Rightarrow2c\left(a+b\right)\ge\left(a+b\right)^2\)

Ta có: \(\left(a+b\right)^2\ge ab\left(a+b\right)^2\)

\(\Leftrightarrow\left(a+b\right)^2\left(1-ab\right)\ge0\)( đúng do (*) ) 

\(\Rightarrow\left(1\right)\)đúng

\(\Rightarrow P-S\ge0\)

\(\Rightarrow P\ge S\)

Ta phải chứng minh \(S\ge0\)

\(\Leftrightarrow\frac{1}{a+b+\frac{1}{a+b}}+a+b+\frac{1}{a+b}\ge\frac{5}{2}\)

\(\Leftrightarrow\frac{a+b}{1+\left(a+b\right)^2}+\frac{1+\left(a+b\right)^2}{a+b}\ge\frac{5}{2}\) (2) 

Đặt \(x=\frac{1+\left(a+b\right)^2}{a+b}\)

Ta có: \(1+\left(a+b\right)^2\ge2\left(a+b\right)\)

\(\Leftrightarrow\left(a+b-1\right)^2\ge0\)( đúng )

\(\Rightarrow x=\frac{1+\left(a+b\right)^2}{a+b}\ge2\)

=> (2) có dạng \(x+\frac{1}{x}\ge\frac{5}{2}\)

\(\Leftrightarrow2x^2-5x+2\ge0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\ge0\)( đúng )

\(\Rightarrow S\ge0\)mà \(P\ge S\)

\(\Rightarrow P\ge0\)

\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{5}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b=1\\ab+bc+ca=1\\ab\left[ab\left(a+b\right)^2+2ab-2\right]=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=c=1;b=0\\b=c=1;a=0\end{cases}}\)

a)

b)

c)

d)

 a)

b)

c)

d)

Giải:

#Học tốt!!!

a) 7 /24

b) 1, 08

c) 17/2

d) 15 /29

a, Ta có  \(\sqrt{25-16}=\sqrt{9}=3\)

\(\sqrt{25}-\sqrt{16}=5-4=1\)

Do 3 > 1 nên \(\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)

a) căn 25 - 16  > căn 25 - căn 16

b)Với $a>b>0$ nên  $\sqrt{a},\sqrt{b},-$ đều xác định

Để so sánh $\sqrt{a}-\sqrt{b}$ và $-$ ta quy về so sánh $\sqrt{a}$ và $-+\sqrt{b}$.

+) $(\sqrt{a}{)}^2=a$.

+) $(-+\sqrt{b}{)}^2=(-{)}^2+2-.\sqrt{b}+(\sqrt{b}{)}^2=a-b+b+2-.\sqrt{b}=a+2-$

$.\sqrt{b}$.

Do $a>b>0$ nên $2-.\sqrt{b}>0$

$\Rightarrow$ $a+2-.\sqrt{b}>a$

$\Rightarrow$ $(-+\sqrt{b}{)}^2>(\sqrt{a}{)}^2$

Do $\sqrt{a},-+\sqrt{b}>0$ 

$\Rightarrow$ $-+\sqrt{b}>\sqrt{a}$

$\iff$ $->\sqrt{a}-\sqrt{b}$ (đpcm)

Vậy $->\sqrt{a}-\sqrt{b}$.

(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)

(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)

(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)

(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)

a) 1/y 

b) - x^2 y 

c) -25x^2 / y^2

d) 4x/5y

a, \(\frac{\sqrt{2}}{\sqrt{18}}=\sqrt{\frac{2}{18}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\)

b, \(\frac{\sqrt{15}}{\sqrt{735}}=\sqrt{\frac{15}{735}}=\sqrt{\frac{1}{49}}=\frac{1}{7}\)

c, \(\frac{\sqrt{12500}}{\sqrt{500}}=\sqrt{\frac{12500}{500}}=\sqrt{\frac{125}{5}}=\sqrt{25}=5\)

d, \(\frac{\sqrt{6^5}}{\sqrt{2^3.3^5}}=\sqrt{\frac{6^5}{2^3.3^5}}=\sqrt{\frac{2^5.3^5}{2^3.3^5}}=\sqrt{2^2}=2\)

a) căn 2 / căn 18 = 1/3

b) căn 15/ căn 735 = 1/7

c) căn 12500 / căn 500 = 5

d) căn 6^5 / 2^3 * 3^5 = 2