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Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
(a+b+c)2=a2+b2+c2
=>2(ab+bc+ac)=0
=>ab+bc+ac=0
=> bc=-ab-ac
=>\(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2-ac-ab+bc}\)=\(\frac{a^2}{\left(a-c\right)\left(a-b\right)}\)
Tuong tu => \(\frac{b^2}{b^2+2ac}=....\)
\(\frac{c^2}{c^2+2ab}=...\)
=> \(\frac{a^2}{a^2+2bc}+....\)=\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)+...
=\(\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
=1
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\)
\(\left(\frac{x}{a}+\frac{y}{b}\right)^2+2\left(\frac{x}{a}+\frac{y}{b}\right)\frac{z}{c}+\left(\frac{z}{c}\right)^2=1\)
\(\left(\frac{x}{a}\right)^2+2\frac{x}{a}\frac{y}{b}+\left(\frac{y}{b}\right)^2+\left(2\frac{x}{a}+2\frac{y}{b}\right)\frac{z}{c}+\left(\frac{z}{c}\right)^2=1\)
\(\frac{x^2}{a^2}+\frac{2xy}{ab}+\frac{y^2}{b^2}+\frac{2xz}{ac}+\frac{2yz}{bc}+\frac{z^2}{c^2}=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\left(\frac{2xy}{ab}+\frac{2xz}{ac}+\frac{2yz}{bc}\right)=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}\left(\frac{c}{z}+\frac{b}{y}+\frac{a}{x}\right)=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}.0=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(ĐPCM\right)\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)
\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
Ta lại có:
\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab+bc-ca}+\frac{b^2}{b^2-ab-bc+ca}+\frac{c^2}{c^2+ab-bc-ca}\)
\(=\frac{a^2}{\left(b-a\right)\left(c-a\right)}+\frac{b^2}{\left(a-b\right)\left(c-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(=-\left(\frac{a^2}{\left(a-b\right)\left(c-a\right)}+\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\left(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)
\(=-\frac{\left(a-b\right)\left(c-b\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Ai có thể giải thích cho mình đoạn a^2/(a^2-ab+bc-ca) đc ko mình cảm ơn
