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Ta co: a3b2=(a2b2)a , a2b3=(a2b2)b => a3b2>a2b3( vi a>b) (1)
b3c2=(b2c2)b , b2c3=(b2c2)c => b3c2>b2c3( vi b>c) (2)
c3a2=(a2c2)c , a3c2=(a2c2)a => c3a2<a3c2 ( vi c<a) (3)
Vi b+c>a ( bdt trong tam giac)
=> dpcm
Bai nay phai xet trong tam giac thi moi dung
\(Q=\frac{a^4}{ab+ca}+\frac{b^4}{ab+bc}+\frac{c^4}{bc+ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}\ge\frac{\frac{\left(a+b+c\right)^2}{3}}{2}\ge\frac{1}{6}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Câu 1:
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(=\dfrac{\left(a+b+c\right)\cdot\left(a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\right)}{2}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]}{2}>=0\)
=>\(a^3+b^3+c^3>=3abc\)
Áp dụng BĐT Cauchy-Schwarz dạng engel:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+b+c+c+a}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cách khác :
Áp dụng BĐT AM-GM cho 2 số dương ta có:
\(\dfrac{a^2}{a+b}+\dfrac{a+b}{4}\ge2\sqrt{\dfrac{a^2\left(a+b\right)}{4\left(a+b\right)}}=a\)
Tương tự: \(\dfrac{b^2}{b+c}+\dfrac{b+c}{4}\ge b;\dfrac{c^2}{c+a}+\dfrac{c+a}{4}\ge c\)
Cộng theo vế ta được:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{a+b+c}{2}\)(đpcm)
Do a , b ,c đối xứng , giả sử a \(\ge b\ge c\Rightarrow\left\{{}\begin{matrix}a^2\ge b^2\ge c^2\\\dfrac{a}{b+c}\ge\dfrac{b}{a+c}\ge\dfrac{c}{a+b}\end{matrix}\right.\)
Áp dụng BĐT Trê - bư -sép ta có :
\(a^2.\dfrac{a}{b+c}+b^2.\dfrac{b}{a+c}+c^2.\dfrac{c}{a+b}\ge\dfrac{a^2+b^2+c^2}{3}.\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)=\dfrac{1}{3}.\dfrac{3}{2}=\dfrac{1}{2}\)Vậy \(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\ge\dfrac{1}{2}\) Dấu bằng xảy ra khi a = b =c = \(\dfrac{1}{\sqrt{3}}\)
Lời giải:
Ta có:
$a^3+b^3-ab(a+b)=(a-b)^2(a+b)\geq 0$ với mọi $a\geq 0; b\geq 0$
$\Rightarrow a^3+b^3\geq ab(a+b)$
Dấu "=" xảy ra khi $a=b$
Có : \(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(a,b\right)=2.3.cos120=-3\)
\(\left(\overrightarrow{a}+\overrightarrow{b}\right)^2=a^2+2\overrightarrow{a}.\overrightarrow{b}+b^2=4-6+9=7\)
\(\Rightarrow\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{7}\)