Câu hỏi t/tự

Ta có \(a+b^2\le\dfrac{a^2+1}{2}+b^2=\dfrac{a^2+2b^2+1}{2}\)

\(\Rightarrow\dfrac{2a^2}{a+b^2}\ge\dfrac{4a^2}{a^2+2b^2+1}=\dfrac{4a^4}{a^4+2b^2a^2+a^2}\). Lập 2 BĐT tương tự rồi áp dụng bất đẳng thức BCS, ta có:

\(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{\left(2a^2+2b^2+2c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\) \(=\dfrac{4\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+3}\)\(=\dfrac{4.3^2}{3^2+3}=3\).

Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\) nên ta có đpcm. ĐTXR \(\Leftrightarrow a=b=c=1\)

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

\(Áp\ dụng\ BĐT\ AM - GM,\ ta\ có: \\\sum\dfrac{1}{a^2+2b^2+3}=\sum\dfrac{1}{(a^2+b^2)+(b^2+1)+2}\le\sum\dfrac{1}{2ab+2b+2} \\=\dfrac{1}{2}\sum\dfrac{1}{ab+b+1}=\dfrac{1}{2}.1=\dfrac{1}{2} \\Đẳng\ thức\ xảy\ ra\ khi\ a=b=c=1.\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{a^3}{2b+3c}+\frac{b^3}{2c+3a}+\frac{c^3}{2a+3b}=\frac{a^4}{2ab+3ac}+\frac{b^4}{2bc+3ba}+\frac{c^4}{2ac+3bc}\)

\(\geq \frac{(a^2+b^2+c^2)^2}{2ab+3ac+2bc+3ba+2ac+3bc}=\frac{(a^2+b^2+c^2)^2}{5(ab+bc+ac)}\)

Theo hệ quả của BĐT AM-GM ta có:

\(a^2+b^2+c^2\geq ab+bc+ac\)

\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)(ab+bc+ac)}{5(ab+bc+ac)}=\frac{a^2+b^2+c^2}{5}\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)

\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)

Tương tự ...

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

\(=\dfrac{1}{2}\left(\dfrac{c}{abc+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{ca.bc+a.bc+bc}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{c}{1+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{c+1+bc}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{c+1+bc}{1+bc+c}\right)=\dfrac{1}{2}\)

\(P_{max}=\dfrac{1}{2}\) khi \(a=b=c=1\)

Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:

\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)

\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)

Tương tự và cộng lại:

\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)

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

\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)

Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z

\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)

Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))

\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))

\(\Rightarrow P\le\dfrac{3}{16}\)

\(ĐTXR\Leftrightarrow a=b=c=1\)