1/ BĐT \(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2\right)+4abc\ge104=\frac{13}{27}\left(a+b+c\right)^3\)
Hay: \(27\left(a+b+c\right)\left(a^2+b^2+c^2\right)+108abc\ge13\left(a+b+c\right)^3\)
\(VT-VP=2\left[6\left\{\Sigma_{cyc}a^3+3abc-\Sigma_{cyc}ab\left(a+b\right)\right\}+\left(a^3+b^3+c^3-3abc\right)\right]\ge0\)
(đúng theo BĐT Schur bậc 3 và Cô si cho 3 số dương)
Đẳng thức xảy ra khi a = b = c = 2
tth_new trả lời nốt luôn đi
đkxđ : \(x,y,z\ge\frac{1}{4}\)
Ta có :
\(x-z=\sqrt{4z-1}-\sqrt{4x-1}=\frac{4\left(z-x\right)}{\sqrt{4z-1}+\sqrt{4x-1}}=-\frac{4\left(x-z\right)}{\sqrt{4z-1}+\sqrt{4x-1}}\)
\(\Rightarrow\left(x-z\right)\left(1+\frac{4}{\sqrt{4z-1}+\sqrt{4x-1}}\right)=0\)
Dễ thấy \(1+\frac{4}{\sqrt{4z-1}+\sqrt{4x-1}}>0\)nên x - z = 0 hay x = z
Tương tự : x = y
Suy ra : x = y = z
Thay vào đầu bài, ta có : \(2x=\sqrt{4x-1}\Rightarrow4x^2=4x-1\Rightarrow x=\frac{1}{2}\)
Vậy x = y = z = \(\frac{1}{2}\)
1 )
Áp dụng BĐT Schur :
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)=\left(6-2a\right)\left(6-2b\right)\left(6-2c\right)\)
\(\Rightarrow abc\ge-216+24\left(ab+bc+ac\right)-8abc\Leftrightarrow3abc\ge8\left(ab+bc+ac\right)-72\)
Do đó
\(VT=3\left(a^2+b^2+c^2\right)+2abc\ge3\left(a^2+b^2+c^2\right)+\frac{16}{3}\left(ab+bc+ac\right)-48\)
\(\Leftrightarrow VT\ge3\left(a+b+c\right)^2-\frac{2}{3}\left(ab+bc+ac\right)-48=60-\frac{2}{3}\left(ab+bc+ac\right)\)
Theo AM - GM : \(ab+bc+ac\le\frac{\left(a+b+c\right)^2}{3}=12\Rightarrow VT\ge52\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=2\)
