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Đặt \(a+b=x,b+c=y,c+a=z\) với \(x,y,z>0\). Ta có:
\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=2\)
\(\Rightarrow\dfrac{1}{x+1}=2-\dfrac{1}{y+1}-\dfrac{1}{z+1}\) \(=1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}\) \(=\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
\(\Rightarrow\dfrac{1}{x+1}\ge2\sqrt{\dfrac{y}{y+1}.\dfrac{z}{z+1}}\)
Tương tự, ta có: \(\dfrac{1}{y+1}\ge2\sqrt{\dfrac{z}{z+1}.\dfrac{x}{x+1}}\) và \(\dfrac{1}{z+1}\ge2\sqrt{\dfrac{x}{x+1}.\dfrac{y}{y+1}}\)
Nhân theo vế 3 BĐT vừa tìm được, ta có:
\(\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}\ge2\sqrt{\dfrac{y}{y+1}.\dfrac{z}{z+1}}.2\sqrt{\dfrac{z}{z+1}.\dfrac{x}{x+1}}.2\sqrt{\dfrac{x}{x+1}.\dfrac{y}{y+1}}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8.\dfrac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Leftrightarrow xyz\le\dfrac{1}{8}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{4}\)
Vậy GTLN của \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\) là \(\dfrac{1}{8}\), xảy ra khi \(a=b=c=\dfrac{1}{4}\)
\(\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{b}\Leftrightarrow ab=bc+ac\Leftrightarrow2ab-2bc-2ac=0\\ \Leftrightarrow\sqrt{a^2+b^2+c^2}=\sqrt{a^2+b^2+c^2+2ab-2bc-2ac}\\ =\sqrt{\left(a+b-c\right)^2}=\left|a+b-c\right|\left(dpcm\right)\)
Áp dụng bđt Svácxơ, ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Áp dụng, thay vào A, ta có:
\(A\le\text{Σ}\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{3}{2}\)
Dấu "="⇔\(a=b=c=1\)
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Sửa \(\le\) thành \(\ge\) nha bạn
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)
Áp dụng BĐT cosi:
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)
\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)
\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)
Cộng VTV:
\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)
Dấu \("="\Leftrightarrow a=b=c=3\)
Với a, b, c là các số dương.
Ta có: \(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}=2\)
\(\Rightarrow\) \(\dfrac{1}{a+b+1}=\left(1-\dfrac{1}{b+c+1}\right)+\left(1-\dfrac{1}{c+a+1}\right) \)
\(=\dfrac{b+c}{b+c+1}+\dfrac{c+a}{c+a+1}\)
\(\ge2\sqrt{\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}>0\) (Bất đẳng thức Cô-si)
Tương tự: \(\dfrac{1}{b+c+1}\ge2\sqrt{\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c+a+1\right)\left(a+b+1\right)}}>0\)
\(\dfrac{1}{c+a+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}>0\)
Nhân vế theo vế ba bất đẳng thức trên, ta được:
\(\dfrac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\) \(\ge\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)
\(\Rightarrow\) \(1\ge8\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\) \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\dfrac{1}{4}\).
Vậy giá trị lớn nhất của tích \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\) bằng \(\dfrac{1}{8}\) khi và chỉ khi \(a=b=c=\dfrac{1}{4}\).