Let A, B, C, and D be positive numbers. Then Lehman’s inequality says

Proof by circuit

This inequality can be proved analytically, but Lehman’s proof is interesting because he uses electrical circuits [1].

Let A, B, C, and D be the resistances of resistors arranges as in the circuit on the left.

Resistors R₁ and R₂ in series have a combined resistance of

and the same resistors in parallel have a combined resistance of

This means the circuit on the left has combined resistance

The resistance of the circuit on the right is

Adding a short cannot increase resistance, so the resistance of the circuit on the right must be the same or lower than the resistance of the one on the left. Therefore

Drawing circuits in LaTeX

I drew the circuits above using the circuitikz package in LaTeX. Here’s the code to draw the circuit on the left.

    documentclass{article}
    
    usepackage{tikz}
    usepackage{circuitikz}
    
    begin{document}
    
    begin{figure}[h!]
      begin{center}
        begin{circuitikz}
          draw (0,0)
          to[R=$B$] (0,2) 
          to[R=$A$] (0,4)
          to[short] (2,4) 
          to[R=$C$] (2,2)
          to[R=$D$] (2,0)
          to[short] (0,0);
          draw (1,4)
          to[short] (1, 5);
          draw (1, 0)
          to[short] (1, -1);
          %draw (0, 2)
          %to[short] (2, 2);
        end{circuitikz}
      end{center}
    end{figure}
    
    end{document}

The code to draw the second circuit removes the %’s commenting out the code that draws the short between the two parallel arms of the circuit.

[1] Alfred Lehman proves a more general result using circuits in SIAM Review, Vol. 4, No. 2 (April 1962) pp. 150–151. Fazlollah Reza gives a purely mathematical proof on pages 151 and 152 of the same journal.

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