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Đặt \(A=abc\left(bc+a^2\right)\left(ac+b^2\right)\left(ab+c^2\right)\)
Do a; b; c > 0 => A > 0
Giả sử \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a+b}{bc+a^2}-\frac{b+c}{ac+b^2}-\frac{c+a}{ab+c^2}\ge0\)
\(\Leftrightarrow\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4a^2c^2-c^4a^2b^2}{A}\ge0\)( tự quy đồng rồi rút gọn nhé, làm chi tiết dài lắm )
\(\Leftrightarrow\frac{2a^4b^4+2b^4c^4+2c^4a^4-2a^4b^2c^2-2b^4a^2c^2-2c^4a^2b^2}{A}\ge0\)
\(\Leftrightarrow\frac{\left(a^2b^2+b^2c^2\right)^2+\left(b^2c^2+c^2a^2\right)^2+\left(c^2a^2+a^2b^2\right)^2}{A}\ge0\)(đúng)
Vậy \(\frac{a+b}{bc+a^2}+\frac{b+c}{ca+b^2}+\frac{c+a}{ab+c^2}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(đpcm)
đặt \(A=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\)
\(\Rightarrow A-3=P=\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ca}\)
áp dụng BĐT cô-si ta có:
\(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca\)
\(\Rightarrow\frac{a^2+b^2}{2}\ge ab;\frac{b^2+c^2}{2}\ge bc;\frac{c^2+a^2}{2}\ge ca\)
\(\Rightarrow1-\frac{a^2+b^2}{2}\le1-ab;1-\frac{b^2+c^2}{2}\le1-bc;1-\frac{c^2+a^2}{2}\le1-ca\)
\(\Rightarrow P\le\frac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{2bc}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\frac{2ca}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\frac{\left(c+a\right)^2}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\right)\)
Áp dụng BĐT Schwarts ta có:
\(\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\)
\(\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\)
\(\frac{\left(c+a\right)^2}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\le\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}\right)=\frac{1}{2}.3=\frac{3}{2}\)
\(\Rightarrow P+3\le\frac{3}{2}+3\)
\(\Rightarrow A\le\frac{9}{2}\)
dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bất đẳng thức cần chứng minh tương đương: \(\frac{1}{ab-1}+\frac{1}{bc-1}+\frac{1}{ca-1}\ge\frac{-9}{2}\)
Theo bất đẳng thức Bunyakovsky dạng phân thức, ta được: \(\frac{1}{ab-1}+\frac{1}{bc-1}+\frac{1}{ca-1}\ge\frac{9}{ab+bc+ca-3}\)
\(\ge\frac{9}{a^2+b^2+c^2-3}=\frac{9}{1-3}=\frac{-9}{2}\left(Q.E.D\right)\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Đặt A= abc(bc+a2)(ac+b2)(ab+c2)
Giả sử 1/a + /b + 1/c - (a+b)/(bc+a2) - (b+c)/(ac+b2) - (c+a)/(ab+c2) >=0
<=> (a4b4+b4c4+c4a4-a4b2c2-b4a2c2-c4a2b2)/A >= 0
<=> (2a4b4+2b4c4+2c4a4-2a4b2c2-2b4a2c2-2c4a2b2)/2A >= 0
<=> (a2b2-b2c2)2+(b2c2-c2a2)2+(c2a2-a2b2)2/2A >= 0 (đúng với mọi a,b,c)
mk chỉ lm theo cách hiểu của mk thôi!nếu ko đúng thì thông cảm nha!
giả sử: \(a\ge b\ge c>0\)(ko mất tính tổng quát)
\(\Rightarrow a^2\ge ac\)\(\Leftrightarrow a^2+bc\ge ac+bc\) (vì b>0;c>0)
\(\Leftrightarrow a^2+bc\ge c\left(a+b\right)\)
\(\Leftrightarrow\frac{a+b}{a^2+bc}\le\frac{1}{c}\) (vì a;b;c>0) (1)
c/m tương tự ta đc: \(\frac{b+c}{ac+b^2}\le\frac{1}{a};\) (2)
\(\frac{c+a}{ab+c^2}\le\frac{1}{b}\) (3)
từ (1),(2),(3)=>đpcm
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
\(\sqrt{\frac{a}{a+bc}}=\frac{a}{\sqrt{a^2+abc}}=\frac{a}{\sqrt{a^2+ab+bc+ca}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Tương tự \(\sqrt{\frac{b}{b+ca}}=\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}};\sqrt{\frac{c}{c+ab}}=\frac{c}{\left(c+a\right)\left(c+b\right)}\)
\(\Rightarrow VT=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(\le\frac{a}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{b}{2}\left(\frac{1}{b+c}+\frac{1}{b+a}\right)+\frac{c}{2}\left(\frac{1}{c+a}+\frac{1}{c+b}\right)\)
\(=\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{c}{a+c}\right)\)
\(=\frac{3}{2}\)
Dấu "=" xảy ra tại \(a=b=c=3\)
1) \(xy\le\frac{\left(x+y\right)^2}{4}\)(cô si) ÁP DỤNG BẤT ĐẲNG THỨC TRÊN với a, b,c>0 TA CÓ
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}+\frac{\left(b+c\right)^2}{4\left(b+c\right)}+\frac{\left(c+a\right)^2}{4\left(c+a\right)}.\)
\(=\frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4}=\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}.\)
2) Với a,b,c >0 .XÉT \(\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}.b}=2a\)(bất đẳng thức cô si)
\(\frac{b^2}{c}+c\ge2\sqrt{\frac{b^2}{c}.c}=2b\)
\(\frac{c^2}{a}+a\ge2\sqrt{\frac{c^2}{a}.a}=2c\)
\(\Rightarrow\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{a}+a\ge2a+2b+2c\)
\(\Leftrightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\)
(đpcm)
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{ab}{2\sqrt{ab}}+\frac{bc}{2\sqrt{bc}}+\frac{ca}{2\sqrt{ca}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
\(VT=\Sigma_{cyc}\frac{1}{a^2-ab+b^2}=\Sigma_{cyc}\frac{abc}{a^2-ab+b^2}=\Sigma_{cyc}\frac{abc}{\left(a-b\right)^2+ab}\)
\(\le\Sigma_{cyc}\frac{abc}{ab}=\Sigma_{cyc}c=a+b+c=VP\)
Đẳng thức xảy ra khi \(a=b=c=1\)
P/s: Mình dùng kí hiệu \(\Sigma_{cyc}\) cho gọn, khi làm bạn tự viết rõ ra.