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![](https://rs.olm.vn/images/avt/0.png?1311)
BĐT tương đương với :
\(3a^4+3b^4+3c^4-\left(a^4+a^3b+a^3c+b^4+ab^3+b^3c+ac^3+bc^3+c^4\right)\ge0\)
\(\Leftrightarrow\left(a^4+b^4-a^3b-ab^3\right)+\left(b^4+c^4-b^3c-bc^3\right)+\left(a^4+c^4-a^3c-ac^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)+\left(b-c\right)^2\left(b^2+bc+c^2\right)+\left(a-c\right)^2\left(a^2+ac+c^2\right)\ge0\)
BĐT cần chứng minh tương đương với:
\(3a^4+3b^4+3c^4\ge a^4+b^4+c^4+ab^3+bc^3+ca^3+a^3b+b^3c+c^3a\)
\(\Leftrightarrow2a^4+2b^4+2c^4-ab^3-bc^3-ca^3-a^3b-b^3c-c^3a\ge0\)
Theo AM - GM ta dễ có:
\(a^4+a^4+a^4+b^4\ge4\sqrt[4]{a^{12}b^4}=4a^3b\)
\(b^4+b^4+b^4+c^4\ge4\sqrt[4]{b^{12}c^4}=4b^3c\)
\(c^4+c^4+c^4+a^4\ge4\sqrt[4]{c^{12}a^4}=4c^3a\)
Cộng vế theo vế ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a^3}{b\left(b+c\right)}+\frac{b}{2}+\frac{b+c}{4}\ge3\sqrt[3]{\frac{a^3}{b\left(b+c\right)}.\frac{b}{2}.\frac{b+c}{4}}=\frac{3}{2}a\)
\(\Leftrightarrow\)\(\frac{a^3}{b\left(b+c\right)}\ge\frac{3}{2}a-\frac{1}{2}b-\frac{1}{4}\left(b+c\right)=\frac{3}{2}a-\frac{3}{4}b-\frac{1}{4}c\)
Tương tự, ta có: \(\frac{b^3}{c\left(c+a\right)}\ge\frac{3}{2}b-\frac{3}{4}c-\frac{1}{4}a;\frac{c^3}{a\left(a+b\right)}\ge\frac{3}{2}c-\frac{3}{4}a-\frac{1}{4}b\)
Cộng theo vế 3 bđt ta được đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
vì a,b,c>=0 =>a=1;b=0;c=0 hoặc a=0;b=1;c=0 hoặc a=0;b=0;c=1
=>a+2b+c>0 mà 1-1=0 => 4(1-a)(1-b)(1-c)=0
=>a+2b+c>=4(1-a)(1-b)(1-c)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Schwarz ta có:
\(\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\ge\frac{\left(2\left(a+b+c\right)\right)^2}{a+b+c}=\frac{4\left(a+b+c\right)^2}{a+b+c}=4\left(a+b+c\right)\)
Dấu ''='' xảy ra bạn tự giải nha.
![](https://rs.olm.vn/images/avt/0.png?1311)
1. a) $(5-2x)^2-16=0$
$=>(5-2x)^2-4^2=0$
$=>(5-2x-4)(5-2x+4)=0$
$=>(1-2x)(9-2x)=0$
\(=>\left[{}\begin{matrix}1-2x=0=>x=0,5\\9-2x=0=>x=4,5\end{matrix}\right.\)
b) $x^2-4x=29$
$=>x^2-4x-29=0$
$=>(x^2-4x+4)-33=0$
$=>(x-2)^2-(\sqrt{33})^2=0$
$=>(x-2-\sqrt{33})(x-2+\sqrt{33})=0$
\(=>\left[{}\begin{matrix}x-2-\sqrt{33}=0=>x=\sqrt{33}+2\\x-2+\sqrt{33}=0=>x=2-\sqrt{33}\end{matrix}\right.\)
Bài 1:
a) \(\left(5-2x\right)^2-16=0\) (1)
\(\Leftrightarrow\left(5-2x\right)^2=16\)
\(\Leftrightarrow5-2x=\pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{1}{2};\dfrac{9}{2}\right\}\)
b) \(x^2-4x=29\) (2)
\(\Leftrightarrow x^2-4x-29=0\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{33}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+2\sqrt{33}}{2}\\x=\dfrac{4-2\sqrt{33}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{2-\sqrt{33};2+\sqrt{33}\right\}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\) (3)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9x^2+18x+9=15\)
\(\Leftrightarrow x^3+27x-27-x^3+27+18x+9=15\)
\(\Leftrightarrow45x+9=15\)
\(\Leftrightarrow45x=15-9\)
\(\Leftrightarrow45x=6\)
\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{2}{15}\right\}\)
d) \(2\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(2x-3\right)+x\left(x^2+8\right)=\left(x+1\right)\left(x^2-x+1\right)\)(4)
\(\Leftrightarrow2\left(x^2-25\right)-\left(2x^2-3x+4x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-\left(2x^2+x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-2x^2-x+6-8x=1\)
\(\Leftrightarrow-44-9x=1\)
\(\Leftrightarrow-9x=1+45\)
\(\Leftrightarrow-9x=45\)
\(\Leftrightarrow x=-5\)
Vậy tập nghiệm phương trình (4) là \(S=\left\{-5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2 ) b )
\(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)
đề sai