\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)

\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).

Đặt biểu thức ở VT là A. Ta có: 

\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).

Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).

Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.

Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).

Đẳng thức xảy ra khi a = b = c = 3.

Vậy...

Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(VT=\dfrac{\sqrt{a^2+a^2+b^2}}{ab}+\dfrac{\sqrt{b^2+b^2+c^2}}{bc}+\dfrac{\sqrt{c^2+a^2+a^2}}{ca}\)

\(VT\ge\dfrac{\sqrt{\dfrac{1}{3}\left(a+a+b\right)^2}}{ab}+\dfrac{\sqrt{\dfrac{1}{3}\left(b+b+c\right)^2}}{bc}+\dfrac{\sqrt{\dfrac{1}{3}\left(c+c+a\right)^2}}{ca}\)

\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)

BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)

Ta có:

\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)

\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)

Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)

\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)

TK: Cho các số thực dương a, b, c thỏa mãn a + b+ c = 3. Chứng minh rằng: \(\sqrt{2a^2+\frac{7}{b^2}}+\sqrt{2b^2+\frac{7}{... - Hoc24

\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)

Áp dụng BĐT Cauchy-Schwarz:

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)

b, \(\frac{a^3}{b+2c}+\frac{b^3}{c+2a}+\frac{c^3}{a+2b}\ge1\)

\(\frac{a^4}{ab+2ac}+\frac{b^4}{bc+2ab}+\frac{c^4}{ac+2bc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac+2ac+2ab+2bc}\)( Bunhia dạng phân thức )

mà \(a^2+b^2+c^2\ge ab+bc+ac\)

\(=\frac{\left(ab+bc+ac\right)^2}{3+2\left(ab+ac+bc\right)}=\frac{9}{3+6}=1\)( đpcm ) 

1.

Điều kiện x \ge \dfrac14x≥41​.

Phương trình tương đương với \left(\sqrt2.\sqrt{2x^2+x+1}-2\right)-\left(\sqrt{4x-1}-1\right)+2x^2+3x-2 = 0(2​.2x2+x+1​−2)−(4x−1​−1)+2x2+3x−2=0 \Leftrightarrow \dfrac{4x^2+2x-2}{\sqrt2.\sqrt{2x^2+x+1}+2} - \dfrac{4x-2}{\sqrt{4x-1}+1} + (x+2)(2x-1) = 0⇔2​.2x2+x+1​+24x2+2x−2​−4x−1​+14x−2​+(x+2)(2x−1)=0\\ \Leftrightarrow (2x-1)\left(\dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2\right) = 0⇔(2x−1)(2​2x2+x+1​+22(x+1)​−4x−1​+12​+x+2)=0

\Leftrightarrow \left[\begin{aligned} & x =\dfrac12\\ & \dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2 = 0\\ \end{aligned}\right.⇔⎣⎢⎢⎢⎡​​x=21​2​2x2+x+1​+22(x+1)​−4x−1​+12​+x+2=0​

Với x \ge \dfrac14x≥41​ ta có:

\dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} > 02​2x2+x+1​+22(x+1)​>0

- \dfrac2{\sqrt{4x-1}+1} \ge -2−4x−1​+12​≥−2

$x+2>2$.

Suy ra \dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2 > 02​2x2+x+1

1.

√2 × √(2x²+x+1)        +      √(4x-1) + 3x-3=0

⇌[√(4x²+2x+2)-2] - [√(4x-1)     -1] + (2x²+3x-2)=0

⇌(4x²+2x-2)/[√(4x²+2x+2)+2] - (4x-2)/[√(4x-1)+1] + (2x-1)(x+2) =0

⇔(2x-1) × [(2x+2)/√(4x²+2x+2+2) - 2/(√4x-1)+1+x+2]=0

Với x≥1/4 thì (2x+2)/(√4x²+2x+2+2)≥0 hoặc x+2>2 hoặc (√4x-1)+1≥1 ⇌ 2/[(√4x-1)+1]≤2

⇒(2x+2)/[(√4x²+2x+2)+2] - 2/[(x-1)+1]+x+2>0-2+2=0

⇌ 2x-1=0⇒x=1/2 

Vậy x=1/2

2.

Áp dụng bất đẳng thức ta có :

Vế trái = a⁴/(ab +2ac)    +   b⁴/(bc+2ab)  + c⁴/(ac+2bc)≥[(a² + b² +c²)²]/[3(ab+bc+ca) =[(a²+b²+c²)²]/9

Ấp dụng bất đẳng thức ta có :

ab+bc+ca≤a²+b²+c² 

Vế trái ≥ [(a²+b²+c²)]/9≥3²/9 =1

⇒ Vế trái ≥1 (đpcm)

Dấu = xảy ra khi a=b=c=1

\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)

Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.

Lời giải vắn tắt:

\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)

( thay \(a^2+b^2+c^2=1\))