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Ta có:
\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)\)
=>Chia 2 vế cho a+b
Ta có:\(a^2+ab+b^2\ge ab\)
=>Trừ 2 vế cho ab \(a^2+b^2\ge0\)
Vì a2 >=0 Với mọi a
b2 >= 0 với mọi b
=> a2+b2>= 0 với mọi a,b
Dấu bằng xảy ra khi:
a2=0 và b2=0
=> a=b=0
Vậy \(a^3+b^3\ge ab\left(a+b\right)\) khi a=b=0
Cách 2 a3+b3>=ab(a+b)
=>a3-a2b +b3-ab2>=)
=> a2(a-b)-b2(a-b)>=0
=>(a-b)2(a+b)>=0 vì a,b dương => a+b>=0
=>Th1:(a-b)2 =0 Th2:a+b=0
=> a-b=0 a=b=0
=>a=b
Vậy a3+b3>= ab(a+b)
\(\frac{a+b+c}{9}\)nha
Đặt \(P=\frac{a^3}{\left(b+2c\right)^2}+\frac{b^3}{\left(c+2a\right)^2}+\frac{c^3}{\left(a+2b\right)^2}\)
Áp dụng bđt AM-GM cho 3 số dương a,b,c ta được:
\(\frac{a^3}{\left(b+2c\right)^2}+\frac{b+2c}{27}+\frac{b+2c}{27}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)^2}.\frac{b+2c}{27}.\frac{b+2c}{27}}=\frac{a}{3}\)
\(\frac{b^3}{\left(c+2a\right)^2}+\frac{c+2a}{27}+\frac{c+2a}{27}\ge3\sqrt[3]{\frac{b^3}{\left(c+2a\right)^2}.\frac{c+2a}{27}.\frac{c+2a}{27}}=\frac{b}{3}\)
\(\frac{c^3}{\left(a+2b\right)^2}+\frac{a+2b}{27}+\frac{a+2b}{27}\ge3\sqrt[3]{\frac{c^3}{\left(a+2b\right)^2}.\frac{a+2b}{27}.\frac{a+2b}{27}}=\frac{c}{3}\)
Cộng từng vế ta được:
\(P+\)\(\frac{6\left(a+b+c\right)}{27}\ge\frac{a+b+c}{3}\)
\(\Rightarrow P\ge\frac{a+b+c}{9}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
\(abc=1\Rightarrow\left(abc\right)^2=a^2b^2c^2=1\Rightarrow a^2=\frac{1}{b^2c^2}\Rightarrow\frac{1}{a^3\left(b+c\right)}=\frac{b^2c^2}{a\left(b+c\right)}=\frac{\left(bc\right)^2}{ab+ac}\)
Chứng minh tương tự ta có: \(\frac{1}{b^3\left(c+a\right)}=\frac{\left(ca\right)^2}{bc+ba};\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{ca+cb}\)
=> \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel: \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{bc+ca+ab+ca+ab+bc}=\frac{ab+bc+ca}{2}\)
Tiếp tục áp dụng bđt Cauchy với 3 số dương ta được: \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3\sqrt[3]{1}}{2}=\frac{3}{2}\)
=> \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{ab+bc+ca}{2}\ge\frac{3}{2}\)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
ta có:\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
=\(\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(a+c\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
>= \(\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\)(BĐT Svaxo)=\(\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
>= \(\frac{3\sqrt[3]{a^2b^2c^2}}{2}\left(BĐTAM-GM\right)=\frac{3}{2}\)(đpcm)
dấu = khi a=b=c=1
1: \(\Leftrightarrow a^5-a^4b+b^5-ab^4>=0\)
\(\Leftrightarrow a^4\left(a-b\right)-b^4\left(a-b\right)>=0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\left(a+b\right)\cdot\left(a^2+b^2\right)>=0\)(luôn đúng khi a,b dương)
\(VT=\frac{\left(a+b+c\right)^2}{9\left(ab+bc+ca\right)}+\frac{ab+bc+ca}{\left(a+b+c\right)^2}+\frac{8\left(a+b+c\right)^2}{9\left(ab+bc+ca\right)}\)
\(VT\ge2\sqrt{\frac{\left(a+b+c\right)^2\left(ab+bc+ca\right)}{9\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\frac{24\left(ab+bc+ca\right)}{9\left(ab+bc+ca\right)}=\frac{10}{3}\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\left(a+b\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)\left(a^3+b^3\right)\)
\(\Leftrightarrow a^5+ab^4+a^4b+b^5\ge a^5+a^2b^3+a^3b^2+b^5\)
\(\Leftrightarrow ab^4+a^4b-a^2b^3-a^3b^2\ge0\)
\(\Leftrightarrow ab\left(a^3+b^3-ab^2-a^2b\right)\ge0\)
\(\Leftrightarrow a^3+b^3-ab^2-a^2b\ge0\)(Do ab > 0)
\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)Luôn đúng do a,b dương
Dấu "='' khi a = b
Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
Áp dụng BĐT AM-GM ta có;
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(\Rightarrow a^2-ab+b^2\ge2ab-ab=ab\)
\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3\ge ab\left(a+b\right)\)
Đẳng thức xảy ra khi \(a=b\)