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Áp dụng bất đẳng thức tam giác :
\(\Rightarrow\left\{\begin{matrix}b+c>a\\c+a>b\\a+b>c\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}b+c+a>2a\\c+a+b>2b\\a+b+c>2c\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}6>2a\\6>2b\\6>2c\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a< 3\\b< 3\\c< 3\end{matrix}\right.\) \(\Rightarrow\left\{\begin{matrix}3-a>0\\3-b>0\\3-c>0\end{matrix}\right.\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{3-a+3-b+3-c}{3}\right)^3\)
\(\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left[\frac{9-\left(a+b+c\right)}{3}\right]^3\)
\(\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{9-6}{3}\right)^3\)
\(\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\le1\)
\(\Rightarrow\left[3\left(3-b\right)-a\left(3-b\right)\right]\left(3-c\right)\le1\)
\(\Rightarrow\left(9-3b-3a+ab\right)\left(3-c\right)\le1\)
\(\Rightarrow3\left(9-3b-3a+ab\right)-c\left(9-3b-3a+ab\right)\le1\)
\(\Rightarrow27-9b-9a+3ab-9c+3bc+3ac-abc\le1\)
\(\Rightarrow27-9b-9a-9c+3ab+3bc+3ac-abc\le1\)
\(\Rightarrow27-9\left(a+b+c\right)+3ab+3bc+3ac-abc\le1\)
Ta có: \(a+b+c=6\)
\(\Rightarrow-27+3ab+3bc+3ac-abc\le1\)
\(\Rightarrow-28+3ab+3bc+3ac\le abc\)
\(\Rightarrow2\left(-28+3ab+3bc+3ac\right)\le2abc\)
\(\Rightarrow2\left(-28+3ab+3bc+3ac\right)+3\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)+2abc\)
\(\Rightarrow-56+6ab+6bc+6ac+3\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)+2abc\)
\(\Rightarrow-56+3\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\le3\left(a^2+b^2+c^2\right)+2abc\)
\(\Rightarrow-56+3\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)+2abc\)
Ta có: \(a+b+c=6\)
\(\Rightarrow-56+3.6^2\le3\left(a^2+b^2+c^2\right)+2abc\)
\(\Rightarrow52\le3\left(a^2+b^2+c^2\right)+2abc\) ( đpcm )
Cách khác:
Áp dụng BĐT Schur:
\(abc\geq (a+b-c)(b+c-a)(c+a-b)=(6-2a)(6-2b)(6-2c)\)
\(\Rightarrow abc\geq -216+24(ab+bc+ac)-8abc\Leftrightarrow 3abc\geq 8(ab+bc+ac)-72\)
Do đó \(\text{VT}=3(a^2+b^2+c^2)+2abc\geq 3(a^2+b^2+c^2)+\frac{16}{3}(ab+bc+ac)-48\)
\(\Leftrightarrow \text{VT}\geq 3(a+b+c)^2-\frac{2}{3}(ab+bc+ac)-48=60-\frac{2}{3}(ab+bc+ac)\)
Theo AM-GM \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=12\Rightarrow \text{VT}\geq 52\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=2$
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 = \(\frac{1}{2}\)
\(a^2\left(b+c\right)+b^2a+b^2c+c^2a+c^2b+2abc=0\)
\(\Leftrightarrow a^2\left(b+c\right)+a\left(b^2+c^2+2bc\right)+bc\left(b+c\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+\left(ab+ac\right)\left(b+c\right)+bc\left(b+c\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(a^2+ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\) đề bài sai
a/ Ta có:
\(\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\le\frac{a+b-c+b+c-a}{2}=b\left(1\right)\)
Tương tự ta có:
\(\hept{\begin{cases}\sqrt{\left(a+b-c\right)\left(c+a-b\right)}\le a\left(2\right)\\\sqrt{\left(b+c-a\right)\left(c+a-b\right)}\le c\left(3\right)\end{cases}}\)
Lấy (1), (2), (3) nhân vế theo vế ta được
\(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le abc\)
\(\frac{1}{c^2\left(a+b\right)}\ge\frac{3}{2};\frac{z^3}{x\left(y+2z\right)}\ge\frac{x+y+z}{3}\)
Tham khảo: Inequality 6