Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

đề sai

1) Tìm GTNN : 

Ta có : \(\frac{x}{y+1}+\frac{y}{x+1}=\frac{x^2}{xy+x}+\frac{y^2}{xy+y}\ge\frac{\left(x+y\right)^2}{2xy+\left(x+y\right)}\ge\frac{1}{\frac{\left(x+y\right)^2}{2}+1}=\frac{1}{\frac{1}{2}+1}=\frac{2}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

2) Áp dụng BĐT Svacxo ta có :

\(\frac{a^2}{1+b}+\frac{b^2}{1+c}+\frac{c^2}{1+a}\ge\frac{\left(a+b+c\right)^2}{3+a+b+c}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

2/ Áp dụng bđt Cô- si cho 2 số dương ta có :

\(\frac{a^2}{1+b}+\frac{1+b}{4}\ge2\sqrt{\frac{a^2}{1+b}\frac{1+b}{4}}=a\)

Tương tự ta có \(\frac{b^2}{1+c}+\frac{1+c}{4}\ge b;\frac{c^2}{1+a}+\frac{1+a}{4}\ge c\)

\(\Rightarrow\frac{a^2}{1+b}+\frac{b^2}{1+c}+\frac{c^2}{1+a}\ge a+b+c-\left(\frac{1+b}{4}+\frac{1+c}{4}+\frac{1+a}{4}\right)\)

\(\Rightarrow\frac{a^2}{1+b}+\frac{b^2}{1+c}+\frac{c^2}{1+a}\ge3-\frac{1}{4}\left(a+b+c\right)-\frac{3}{4}=3-\frac{1}{4}.3-\frac{3}{4}=\frac{3}{2}\)

Dấu "=" xảy ra <=> a=b=c=1 

a)Áp dụng Bđt cô si, ta có:

\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự ta có:

\(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)

Cộng 3 vế của bđt lại ta có:

\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}=a+b+c-\frac{ab+bc+ac}{2}\ge\frac{3}{2}\)

dấu = khi a=b=c=1

a)Từ \(a+b+c\ge ab+bc+ca\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) *đúng*

Khi \(a=b=c\)

b)Áp dụng BĐT AM-GM ta có: 

\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự rồi cộng theo vế :

\(M\ge3-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)

Khi \(a=b=c=1\)

Ta sẽ sử dụng đánh giá \(x^3+\frac{1}{x^3}\ge\frac{1}{\left(1+9^3\right)^2}\left(x+\frac{81}{x}\right)^3\)

Dấu "=" xảy ra <=> x=\(\frac{1}{3}\)

Sử dụng đánh giá trên ta có: \(\hept{\begin{cases}\sqrt[3]{a^3+\frac{1}{a^3}}\ge\frac{1}{\sqrt[3]{\left(1+9^3\right)^2}}\left(a+\frac{81}{a}\right)\\\sqrt[3]{b^3+\frac{1}{b^3}}\ge\frac{1}{\sqrt[3]{\left(1+9^3\right)^2}}\left(b+\frac{81}{b}\right)\\\sqrt[3]{c^3+\frac{1}{c^3}}\ge\frac{1}{\sqrt[3]{\left(1+9^3\right)^2}}\left(c+\frac{81}{c}\right)\end{cases}}\)

Cộng theo vế ta được \(P=\sqrt[3]{a^3+\frac{1}{a^3}}+\sqrt[3]{b^3+\frac{1}{b^3}}+\sqrt[3]{c^3+\frac{1}{c^3}}\ge\frac{1}{\sqrt[3]{\left(1+9^3\right)^2}}\left(a+b+c+\frac{81}{a}+\frac{81}{b}+\frac{81}{c}\right)\)

Ta lại có: \(a+b+c+\frac{81}{a}+\frac{81}{b}+\frac{81}{c}\ge a+b+c+\frac{729}{a+b+c}=a+b+c+\frac{1}{a+b+c}+\frac{729}{a+b+c}\)

\(\ge2+728=730\)

=> \(P\ge\frac{730}{\sqrt[3]{\left(1+9^3\right)^2}}=\sqrt[3]{730}\)

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)

Hey Hải Nhật, mk có bảo bạn giải đâu ạ? Lời giải này thì mk biết lâu r, (chép trong tài liệu), nhưng mình hỏi cách tìm bđt phụ kia cơ mà

\(P=1\sqrt{a-1}+1\sqrt{b-2}+1\sqrt{c-3}\le\dfrac{1}{2}\left(1+a-1+1+b-2+1+c-3\right)=3\)

\(P_{max}=3\) khi \(\left(a;b;c\right)=\left(2;3;4\right)\)

\(P^2=a+b+c-6+2\left(\sqrt{\left(a-1\right)\left(b-2\right)}+\sqrt{\left(a-1\right)\left(c-3\right)}+\sqrt{\left(b-2\right)\left(c-3\right)}\right)\)

\(P^2\ge a+b+c-6=3\)

\(P\ge\sqrt{3}\)

\(P_{min}=\sqrt{3}\) khi \(\left(a;b;c\right)=\left(1;2;6\right);\left(1;5;3\right);\left(4;2;3\right)\)

Kìa có GV dễ xương box toán giải rèo kìa 😂

Ta có: \(1=a^2+b^2+c^2\ge ab+bc+ca\).

\(P=\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b}=\dfrac{a^4}{ab+2ca}+\dfrac{b^4}{bc+2ab}+\dfrac{c^4}{ca+2bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3\left(ab+bc+ca\right)}=\dfrac{1}{3\left(ab+bc+ca\right)}\ge\dfrac{1}{3}\)

Dấu \(=\) xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\).