Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn

Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng

em cũng nghĩ thế mới dùng đc BDT AM-GM 3 số đúng ko thầy :)

Theo giả thiết kết hợp sử dụng BĐT AM - GM có:

\(\left(a+b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-\left[c\left(a+b\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)

\(\le\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-2\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\left[\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\right]^2\)

Suy ra \(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\ge2\Leftrightarrow\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}\ge3\)

\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge7\)

Khi đó, sử dụng BĐT Cauchy - Schwarz ta có:

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

\(=\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+1\right)^2=\left[\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-1\right]^2\ge\left(7^2-1\right)^2=2304\)

Đẳng thức xảy ra khi và chỉ khi \(ab=c^2\) và \(\dfrac{a}{b}+\dfrac{b}{a}=7\)

(a+b-c)(1/a+1/b-c)=(a+b)(1/a+1/b)+1-[c(a+b)+c(1/a+1/b)]<=(a+b)(1/a+1/b)+1-2căn (a+b)(1/a+1/b)

=[(căn (a+b)(1/a+1/b))-1]^2

=>\(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1>=2\)

=>\(\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}>=3\)

=>a/b+b/a>=7

(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)>=[căn ((a^4+b^4)(1/a^4+1/b^4))+1]^2

=(a^2/b^2+b^2/a^2+1)^2=[(a/b+b/a)^2-1]^2>=(7^2-1)^2=2304

=>ĐPCM

Bài này đã có ở đây:

Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24

\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)

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

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

Áp dụng BĐT quen thuộc:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)

Ta chỉ cần chứng minh:

\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)

Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)

Nên ta chỉ cần chứng minh:

\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)

\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)

`sqrta+sqrtb+sqrtc=2`

`<=>(sqrta+sqrtb+sqrtc)^2=4`

`<=>a+b+c+2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4`

`<=>2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4-(a+b+c)=4-2-2`

`<=>sqrt{ab}+sqrt{bc}+sqrt{ca}=1`

`=>a+1=a+sqrt{ab}+sqrt{bc}+sqrt{ca}=sqrta(sqrta+sqrtb)+sqrtc(sqrta+sqrtb)=(sqrta+sqrtb)(sqrta+sqrtc)`

Tương tự:`b+1=(sqrtb+sqrta)(sqrtb+sqrtc)`

`c+1=(sqrtc+sqrta)(sqrtc+sqrtb)`

`=>VT=sqrta/((sqrta+sqrtb)(sqrta+sqrtc))+sqrtb/((sqrtb+sqrta)(sqrtb+sqrtc))+sqrtc/((sqrtc+sqrta)(sqrtc+sqrtb))`

`=>VT=(sqrta(sqrtb+sqrtc)+sqrtb(sqrtc+sqrta)+sqrtc(sqrta+sqrtb))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(sqrt{ab}+sqrt{ac}+sqrt{bc}+sqrt{ab}+sqrt{ac}+sqrt{bc})/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(2(sqrt{ab}+sqrt{bc}+sqrt{ca}))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/\sqrt{[(sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta)]^2}`

`=2/\sqrt{(sqrta+sqrtb)(sqrta+sqrtc)(sqrtb+sqrta)(sqrtb+sqrtc)(sqrtc+sqrta)(sqrtc+sqrtb)}`

`=2/\sqrt{(1+a)(1+b)(1+c)}=>đpcm`