Áp dụng BĐT Cô - si cho 3 số không âm:

\(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{a^3}{b^3}}+1\ge3\sqrt[3]{\sqrt{\frac{a^6}{b^6}}}=\frac{3a}{b}\)

\(\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{b^3}{c^3}}+1\ge3\sqrt[3]{\sqrt{\frac{b^6}{c^6}}}=\frac{3b}{c}\)

\(\sqrt{\frac{c^3}{a^3}}+\sqrt{\frac{c^3}{a^3}}+1\ge3\sqrt[3]{\sqrt{\frac{c^6}{a^6}}}=\frac{3c}{a}\)

Cộng vế theo vế ,ta được:

\(2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)\(+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)\(+3\)

\(\Rightarrow2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

Vậy \(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)(đpcm)

Trâu bò chút!

Đặt \(\sqrt{\frac{a}{b}}=x;\sqrt{\frac{b}{c}}=y;\sqrt{\frac{c}{a}}=z\Rightarrow xyz=1\)

BĐT quy về chứng minh: \(x^3+y^3+z^3\ge x^2+y^2+z^2\)

Để ý rằng: \(x^3=\frac{\left(x-1\right)^2\left(2x+1\right)}{2}+\frac{3}{2}x^2-\frac{1}{2}\)

Từ đó ta có:  \(VT-VP=\Sigma_{cyc}\frac{\left(x-1\right)^2\left(2x+1\right)}{2}+\frac{1}{2}\left(\Sigma x^2-3\right)\)

\(\ge\Sigma_{cyc}\frac{\left(x-1\right)^2\left(2x+1\right)}{2}\ge0\)

P/s: Nếu thích troll người thì thế ngược lại các biến đã đặt ta tìm được:

\(VT-VP\ge\Sigma_{cyc}\frac{\left(a-b\right)^2\left(2\sqrt{a}+\sqrt{b}\right)}{2b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)^2}\ge0\)

Do \(a+b+c=1\)  nên :

\(VT=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\frac{ca}{b\left(a+b+c\right)+ac}}\)

\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)

Áp dụng BĐT AM - GM :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)

\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)

Cộng theo vế :
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\left(đpcm\right)\)

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

Chúc bạn học tốt !!!

Ta có: 

\(AB+BC=AC\sqrt{3}\)

=> \(\frac{AC}{BC}\sqrt{3}-\frac{AB}{BC}=1\)

=> \(\sqrt{3}\cos\widehat{C}-\sin\widehat{C}=1\)

=> \(\sqrt{3}\cos\widehat{C}-1=\sin\widehat{C}\)

Mặt khác: \(\sin^2\widehat{C}+\cos^2\widehat{C}=1\)

<=> \(\left(\sqrt{3}\cos\widehat{C}-1\right)^2+\cos^2\widehat{C}=1\)

<=> \(4\cos^2\widehat{C}-2\sqrt{3}\cos\widehat{C}=0\)( \(0^o< \widehat{C}< 90^o\))

<=> \(2\cos\widehat{C}-\sqrt{3}=0\)

<=> \(\cos\widehat{C}=\frac{\sqrt{3}}{2}\)

<=> \(\widehat{C}=30^o\)=> \(\widehat{B}=60^o\)

em cảm ơn cô nhiều lắm ạ <3

Đặt: \(\sqrt{x}=t\)( \(t\ge0;t\ne1\)) => \(A\ne0\)

Ta có: \(A=\frac{t-1}{t^2+t+1}\)

<=> \(At^2+At+A=t-1\)

<=> \(At^2+\left(A-1\right)t+\left(A+1\right)=0\) (1)

(1) có nghiệm <=> \(\Delta\ge0\)<=> \(-3A^2-6A+1\ge0\)<=> \(-1-\frac{2}{\sqrt{3}}\le A\le-1+\frac{2}{\sqrt{3}}\)

Theo đề ra A thuộc Z ; A khác 0

=> A \(\in\){ - 2; -1 }

+) Với A = - 2  thế vào (1) ta có: \(-2t^2-3t-1=0\) <=> \(\orbr{\begin{cases}t=-1\left(loai\right)\\t=-\frac{1}{2}\left(loai\right)\end{cases}}\)

+) Với A = -1 thế vào (1) ta có: \(-t^2-2t=0\)<=> \(\orbr{\begin{cases}t=0\left(tm\right)\\t=-2\left(loai\right)\end{cases}}\)

Với t = 0 ta có: \(\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

Vậy x = 0 ; A = -1

E cảm ơn  cô