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\(ab+bc+ca\le1\)
\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)
\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)
\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)
Đề bài sai, bạn kiểm tra lại điều kiện \(a^2+b^2+c^2=1\)
\(VT=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\)
\(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
\(VT=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{2}{\left(a+1\right)^2}+\dfrac{2}{\left(b+1\right)^2}+\dfrac{2}{\left(c+1\right)^2}\)
Mặt khác:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1.1\right)^2}\ge\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{a}{b}\right)}+\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{b}{a}\right)}=\dfrac{1}{1+ab}\)
Do đó:
\(VT\ge\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)
\(VT\ge\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{1}{1+\dfrac{1}{c}}+\dfrac{1}{1+\dfrac{1}{a}}+\dfrac{1}{1+\dfrac{1}{b}}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
cho em hỏi một tí ạ
Chộ \(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1.1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(1+\dfrac{a}{b}\right)}+\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{b}{a}\right)}\)
áp dụng công thức gì đây ạ
\(\frac{\left(1+a\right)^2+b^2+5}{ab+a+4}=\frac{a^2+b^2+2a+6}{ab+a+4}\ge\frac{2ab+2a+6}{ab+a+4}=2-\frac{2}{ab+a+1+3}\ge2-\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{3}\right)\)
Tương tự: \(\frac{\left(1+b\right)^2+c^2+5}{bc+b+4}\ge2-\frac{1}{2}\left(\frac{1}{bc+b+1}+\frac{1}{3}\right)\) ; \(\frac{\left(1+c\right)^2+c^2+5}{ac+c+4}\ge2-\frac{1}{2}\left(\frac{1}{ac+c+1}+\frac{1}{3}\right)\)
Cộng vế với vế:
\(P\ge\frac{11}{2}-\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{11}{2}-\frac{1}{2}=5\)
\(P_{min}=5\) khi \(a=b=c=1\)
\(a^5+b^2+ab+6\ge3a^2b+6\)
\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)
\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)
\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)
\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
\(\Rightarrow P\le\sqrt{1}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
điện thoại cùi nên chụp hơi mờ, đề này còn thiếu a,,bc>0
\(A=3\left(ab+bc+ca\right)+\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{4}\left(b-c\right)^2+\dfrac{1}{8}\left(c-a\right)^2\\ =3\left(ab+bc+ca\right)+\dfrac{\left(a-b\right)^2}{2}+\dfrac{\left(b-c\right)^2}{4}+\dfrac{\left(c-a\right)^2}{8}\)
Áp dụng BDT: Cô-si dạng Engel:
\(\Rightarrow A=3\left(ab+bc+ca\right)+\dfrac{\left(a-b\right)^2}{2}+\dfrac{\left(b-c\right)^2}{4}+\dfrac{\left(c-a\right)^2}{8}\ge3\left(ab+bc+ca\right)+\dfrac{\left(a-b+b-c+c-a\right)^2}{2+4+8}=3\left(ab+bc+ca\right)\left(1\right)\)
\(\text{Ta lại có: }ab+bc+ac\le a^2+b^2+c^2\\ \Leftrightarrow ab+bc+ac+2\left(ab+bc+ac\right)\le a^2+b^2+c^2+2\left(ab+bc+ac\right)\\ \Leftrightarrow3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2=3^2=9\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow A\le9\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}a=b=c\\a+b+c=3\\\dfrac{a-b}{2}+\dfrac{b-c}{4}+\dfrac{c-a}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\Leftrightarrow a=b=c=1\)
Vậy \(A_{Max}=9\) khi \(a=b=c=1\)
Ta có: \(\dfrac{1}{4-\sqrt{ab}}\le\dfrac{1}{4-\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(\left(a^2+b^2;b^2+c^2;c^2+a^2\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x+y+z=6\\x;y;z>0\end{matrix}\right.\)
Làm nốt :v