Lời giải:
Áp dụng BĐT AM-GM cho các số dương:
\((a+b-c)(b+c-a)\leq \left(\frac{a+b-c+b+c-a}{2}\right)^2=b^2\)

\((a+b-c)(c+a-b)\leq \left(\frac{a+b-c+c+a-b}{2}\right)^2=a^2\)

\((b+c-a)(c+a-b)\leq \left(\frac{b+c-a+c+a-b}{2}\right)^2=c^2\)

Nhân theo vế và rút gọn :

\(\Rightarrow (a+b-c)(b+c-a)(c+a-b)\leq abc\)

\(\Leftrightarrow (6-2c)(6-2a)(6-2b)\leq abc\) (do $a+b+c=6$)

\(\Leftrightarrow 8[27-9(a+b+c)+3(ab+bc+ac)-abc]\leq abc\)

\(\Leftrightarrow 8(-27+3(ab+bc+ac)-abc)\leq abc\)

\(\Leftrightarrow abc\geq \frac{8}{3}(ab+bc+ac)-24\)

Do đó:

\(3(a^2+b^2+c^2)+2abc\geq 3(a^2+b^2+c^2)+\frac{16}{3}(ab+bc+ac)-48\)

\(=3(a+b+c)^2-\frac{2}{3}(ab+bc+ac)-48=60-\frac{2}{3}(ab+bc+ac)\)

Mà theo hệ quả của BĐT AM-GM \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=12\)

\(\Rightarrow 3(a^2+b^2+c^2)+2abc\geq 60-\frac{2}{3}.12=52\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

Từ gt suy ra a < b + c nên 2a < a + b + c = 2

\(\Rightarrow a< 1\).

Chứng minh tương tự: \(b< 1;c< 1\).

Do đó \(\left(a-1\right)\left(b-1\right)\left(c-1\right)< 0\Leftrightarrow abc< ab+bc+ca-1\) (Do a + b + c = 2)

\(\Rightarrow a^2+b^2+c^2+2abc< a^2+b^2+c^2+2\left(ab+bc+ca-1\right)=\left(a+b+c\right)^2-2=2\) (đpcm).

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 $a=b=c=2$

a, b, c là độ dài 3 cạnh của tgiác nên ta có: b+c > a => ab+ac > a² 
tương tự: bc+ab > b²; ca+bc > c² 
cộng lại: 2ab+2bc+2ca > a²+b²+c² (*) 

gthiết: 4 = (a+b+c)² = a²+b²+c² + 2ab+2bc+2ca > a²+b²+c² + a²+b²+c² {ad (*)} 
=> 2 > a²+b²+c² (đpcm) 

đúng nha

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 \(a=b=c=2\)