Sr chụy nha, em chưa học tới ~ :]]]

bdt tương đương với  \(a^2+b^2+c^2+d^2+2ac+2bd\le a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\)

\(\Leftrightarrow2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge2\left(ac+bd\right)\)

\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(b^2+d^2\right)}\ge ac+bd\)

neu ac+bd \(\le0\) thi bdt can duoc cm 

neu ac+bd \(\ge0\) thi \(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge a^2c^2+b^2d^2+2abcd\)

                \(\Leftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2\ge a^2c^2+b^2d^2+2abcd\)

 \(\Leftrightarrow b^2c^2+a^2d^2-2abcd\ge0\Leftrightarrow\left(bc-ad\right)^2\ge0\left(dpcm\right)\)

Giả sử BĐT đúng , Bình phương 2 vế đc

\(a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge a^2+b^2+c^2+d^2+2\left(ac+bd\right)\)

\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\).Bình phương 2 vế đc

\(\Leftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2\ge a^2c^2+b^2d^2+2abcd\)

\(\Leftrightarrow a^2d^2+b^2c^2\ge2abcd\Leftrightarrow\left(ad-bc\right)^2\ge0\)(luôn đúng)

Vậy BĐT luôn đúng mà bạn ghi sai dấu

có thiếu ĐK nào k bạn ?

áp dụng BĐT cauchy :

\(\dfrac{b}{\left(a+\sqrt{b}\right)^2}+\dfrac{d}{\left(c+\sqrt{d}\right)^2}\ge2\sqrt{\dfrac{bd}{\left(a+\sqrt{b}\right)^2\left(c+\sqrt{d}\right)^2}}=\dfrac{2\sqrt{bd}}{\left(a+\sqrt{b}\right)\left(c+\sqrt{d}\right)}\)

việc còn lại cần chứng minh \(\left(a+\sqrt{b}\right)\left(c+\sqrt{d}\right)\le2\left(ac+\sqrt{bd}\right)\)(đúng theo BĐT chebyshev)(không mất tính tổng quát giả sừ \(a\le\sqrt{b};c\le\sqrt{d}\))

dấu = xảy ra khi \(a=\sqrt{b};c=\sqrt{d}\)

b: \(A=\dfrac{x^2+4+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\dfrac{1}{\sqrt{x^2+4}}>=2\sqrt{\sqrt{x^2+4}\cdot\dfrac{1}{\sqrt{x^2+4}}}=2\)

a: =>ab+ad+bc+cd>=ab+cd+2căn abcd

=>ad+cb-2căn abcd>=0

=>(căn ad-căn cb)^2>=0(luôn đúng)

a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)

\(=-4\sqrt{5}+15\sqrt{2}\)

b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)

\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)

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

\(=8\sqrt{3}+2\sqrt{2}-4\)

c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)

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

=9-3

=6

d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

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

=4

Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\) 

Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

 Dạ em cám ơn nhiều lắm ạ

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

\(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)