Ta có: \(a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

a) \(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2\ge4abcd\)

b) \(a^2+1\ge2a,b^2+1\ge2b,c^2+1\ge2c\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge8abc\)

c) \(a^2+4\ge4a,b^2+4\ge4b,c^2+4\ge4c,d^2+4\ge4d\)

\(\Rightarrow\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)

a) \(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2=2\left[\left(ab\right)^2+\left(cd\right)^2\right]\ge2\cdot2abcd=4abcd\)

b) \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2a\cdot2b\cdot2c=8abc\)

c) \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge4a\cdot4b\cdot4c\cdot4d=256abcd\)

Câu 1 cần bổ sung thêm điều kiện $a,b,c$ là 3 cạnh của tam giác, tức là đảm bảo mẫu các phân thức vế trái luôn dương.

Nếu không, BĐT sai trong TH $(a,b,c)=(3,2,10)$

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

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

\(=\frac{(a^2+b^2+c^2)^2}{2(ab+bc+ac)-(a^2+b^2+c^2)}(1)\)

Mà theo BĐT AM-GM ta thấy: $ab+bc+ac\leq a^2+b^2+c^2$

$\Rightarrow 2(ab+bc+ac)-(a^2+b^2+c^2)\leq a^2+b^2+c^2(2)$

Từ $(1);(2)\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2}=a^2+b^2+c^2$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

1)ĐK:\(x\in\left[-3;\frac{6}{5}\right]\)

pt\(\Leftrightarrow3\left(x^2-x+2\right)-3\left[\sqrt{6-5x}-\left(x-2\right)\right]+\left[3\sqrt{x+3}-\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x^2-x+2\right)\left(\frac{3}{\sqrt{6-5x}+x-2}+\frac{1}{3\sqrt{x+3}+x+5}+3\right)=0\)

\(\Leftrightarrow x^2\)-x+2=0(do(...)>0)

\(\Leftrightarrow x=-2\)hoặc \(x=1\)(t/m)

ÁD BĐT Bunhiacopxki:

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

Lại có:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

\(=\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\)

\(\Rightarrow VT\ge\left(\frac{3}{2}\right)^2\)=\(\frac{9}{4}\)(đpcm)

Dấu''='' xảy ra\(\Leftrightarrow a=b=c=\frac{1}{3}\)

BĐT \(\Leftrightarrow\left(\sqrt{ab}+\sqrt{cd}\right)^2\le\left(a+d\right)\left(b+c\right)\)

\(\Leftrightarrow ab+2\sqrt{abcd}+cd\le ab+ac+bd+dc\)

\(\Leftrightarrow2\sqrt{abcd}\le ac+bd\)

\(\Leftrightarrow0\le\left(\sqrt{ac}-\sqrt{bd}\right)^2\) ( luôn đúng )

Dấu "=" xảy ra khi \(\sqrt{ac}=\sqrt{bd}\Leftrightarrow ac=bd\Leftrightarrow\frac{a}{b}=\frac{d}{c}\)

đk đâu?

3: =>a^2c^2+a^2d^2+b^2c^2+b^2d^2>=a^2c^2+2abcd+b^2d^2

=>a^2d^2-2abcd+b^2c^2>=0

=>(ad-bc)^2>=0(luôn đúng)

Áp dụng liên tiếp AM-GM và Cauchy-Schwarz ta có:

\(\begin{align*} \dfrac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}&\ge \dfrac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left (a^2+b^2 \right )}}\\ &=\dfrac{a^2+ab+1}{\sqrt{a^2+ab+1}}\\ &=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\\ &=\dfrac{1}{\sqrt{5}}\sqrt{\left ( \dfrac{9}{4}+\dfrac{3}{4}+1+1 \right )\left [\left ( a+\dfrac{b}{2} \right )^2+\dfrac{3b^2}{4}+a^2+c^2 \right ]}\\ &\ge \dfrac{1}{\sqrt{5}}\left [ \dfrac{3}{2}\left (a+\dfrac{b}{2} \right )+\dfrac{3}{4}b+a+c \right ]\\ &=\dfrac{1}{\sqrt{5}}\left ( \dfrac{5}{2}a+\dfrac{3}{2}b+c \right ) \end{align*}\)

Chứng minh tương tự, cộng lại ta có đpcm.

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

bài này cuốn hút thật, lâu lắm ms thấy . xí bài này nhé nghĩ đã lát quay lại làm