Nice proof, nhưng đã quy đồng là phải thế này :v

\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)

\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)

\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)

\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)

Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:

\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)

Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)

Áp dụng BĐT này ta có:

\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)

Cộng theo vế 3 BĐT trên ta có:

\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)

\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)

\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)

\(E=\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}=\dfrac{a+\sqrt{a}+2+\sqrt{a}-2}{a-4}=\dfrac{a+2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-2}\)

ko biet

b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)

\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)

\(=\dfrac{a}{a-b}\)

khúc \(\dfrac{a}{a-b}\) sai nhé

\(=\dfrac{a-b}{a-b}=1\)

Câu a : \(VT=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

\(=\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}}+\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{2\left(2-\sqrt{3}\right)}}\)

\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)

\(=\sqrt{\dfrac{3-2\sqrt{3}+1}{3+2\sqrt{3}+1}}+\sqrt{\dfrac{3+2\sqrt{3}+1}{3-2\sqrt{3}+1}}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)

\(=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\dfrac{3-2\sqrt{3}+1+3+2\sqrt{3}+1}{3-1}\)

\(=\dfrac{8}{2}=4\) ( đpcm )

Câu c : \(VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\) ( đpcm )

Bài 1:

Áp dụng BĐT AM-GM ta có:

$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$

Cộng theo vế và thu gọn:

$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$

Ta có đpcm.

Bài 2:

$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$

$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$

$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$

Cộng theo vế và rút gọn thu được:

$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$ 

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

a) Ta có: \(A=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^2-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^4-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^8-1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{16}-1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{32}-1\right]\)

\(=65535\sqrt{2}+65535\)

b) Ta có: \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2020}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)

\(=\sqrt{2020}-1\)

\(=2\sqrt{505}-1\)

c) Ta có: \(C^3=26+15\sqrt{3}+26-15\sqrt{3}+3\cdot\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\cdot\left(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\right)\)

\(\Leftrightarrow C^3=52+3\cdot C\)

\(\Leftrightarrow C^3-3\cdot C-52=0\)

\(\Leftrightarrow C^3-4C^2+4C^2-16C+13C-52=0\)

\(\Leftrightarrow C^2\left(C-4\right)+4C\left(C-4\right)+13\left(C-4\right)=0\)

\(\Leftrightarrow\left(C-4\right)\left(C^2+4C+13\right)=0\)

mà \(C^2+4C+13>0\)

nên C-4=0

hay C=4

a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)

b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)

c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)