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Cần CM: \(\frac{a}{\left(1-a\right)^3}\ge\frac{135}{16}a-\frac{27}{16}\)\(\left(0< a< 1\right)\)

thaajt vậy, bđt \(\Leftrightarrow\)\(\left(a-\frac{1}{3}\right)^2\left(15a^2-38a+27\right)\ge0\) đúng 

\(\Sigma\frac{a}{\left(b+c\right)^3}=\Sigma\frac{a}{\left(1-a\right)^3}\ge\frac{135}{16}\left(a+b+c\right)-\frac{81}{16}=\frac{27}{8}\)

dấu "=" xảy ra khi a=b=c=1 

à nhầm, \(a=b=c=\frac{1}{3}\)

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

\(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left(a^2+b^2\right)}}\)

   \(=\frac{a^2+ab+1}{\sqrt{a^2+ab+1}}\)

\(=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\)

\(=\frac{1}{\sqrt{5}}\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+a^2+c^2\right]}\)

\(\ge\frac{1}{\sqrt{5}}\left[\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{4}b+a+c\right]\)

\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)

Chứng minh tương tự và công lại ta có đpcm 

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

\(\sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=3\sqrt{9a^2}=3.3.a=9a\) ( vì \(a\ge0\) )

triểu khai bt : \(\sqrt{3\cdot a\cdot3\cdot9\cdot a}\)= 9a

Ta có : \(K=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}\)

\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}\)

\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}=\frac{3\sqrt{2}+\sqrt{6}}{6}\)

Dăm 3 vài phương pháp sos:)

\(P=\Sigma_{cyc}\frac{\left(x-y\right)^2\left(x+2y\right)}{y^2}+\left(x+y+z\right)\ge x+y+z=3\)

Đẳng thức xảy ra khi x = y = z = 1

ĐKXĐ : ....

PT \(\Leftrightarrow\sqrt{-x^2+4x-3}-1+\sqrt{-2x^2+8x+1}-3=x\left(x^2-4x+4\right)\)

\(\Leftrightarrow\frac{-x^2+4x-4}{\sqrt{-x^2+4x-3}+1}+\frac{-2x^2+8x-8}{\sqrt{-2x^2+8x+1}+3}=x\left(x-2\right)^2\)

\(\Leftrightarrow\frac{\left(x-2\right)^2}{\sqrt{-x^2+4x-3}+1}+\frac{2\left(x-2\right)^2}{\sqrt{-2x^2+8x+1}+3}+x\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{-x^2+4x-3}+1}+\frac{2}{\sqrt{-2x^2+8x+1}+3}+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^2=0\\\frac{1}{\sqrt{-x^2+4x-3}+1}+\frac{2}{\sqrt{-2x^2+8x+1}+3}+x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{\sqrt{-x^2+4x-3}+1}+\frac{2}{\sqrt{-2x^2+8x+1}+3}+x>0\left(loai\right)\end{cases}}\)

Bạn Thanh Tùng DZ ơi sao trường hợp 2 lại loại vậy

Chưa có điều kiện của x mà

\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\left(true!!\right)\)

Dấu "=" xảy ra tại a=b=1

Xét hiệu \(A=\left(a^2+b^2+1\right)-\left(ab+a+b\right)\)

\(=a^2+b^2+1-ab-a-b\)

\(\Rightarrow2A=2a^2+2b^2+2-2ab-2a-2b\)

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

\(=\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)

\(\Rightarrow2A\ge0\Leftrightarrow A\ge0\)

Vậy \(a^2+b^2+1\ge ab+a+b\left(đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\a-1=0\\b-1=0\end{cases}}\Leftrightarrow a=b=1\)

Ta có : \(ab+bc+ca=2abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có :

\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{12}\)

Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)